JEE Advanced PYQ System of Particles and Rotational Motion

Q 1 :

A disc is rolling without slipping with angular velocity $ω$ . P and Q are two points equidistant from the centre C. The order of magnitude of velocity is [2004]

[IMAGE 207]

$v_{Q} > v_{C} > v_{P}$
$v_{P} > v_{C} > v_{Q}$
$v_{P} = v_{C}, v_{Q} = \frac{v_{C}}{2}$
$v_{P} < v_{C} > v_{Q}$

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(1)

In pure rolling, the point of contact is the instantaneous centre of rotation of all the particles of the disc.

On applying $v = r ω$

We find $ω$ is same for all the particles, then $v \propto r .$

$r_{Q} > r_{C} > r_{P} \Rightarrow v_{Q} > v_{C} > v_{P}$

Q 2 :

A small roller of diameter 20 cm has an axle of diameter 10 cm (see figure below on the left). It is on a horizontal floor and a meter scale is positioned horizontally on its axle with one edge of the scale on top of the axle (see figure on the right). The scale is now pushed slowly on the axle so that it moves without slipping on the axle, and the roller starts rolling without slipping. After the roller has moved 50 cm, the position of the scale will look like (figures are schematic and not drawn to scale) [2020]

[IMAGE 208]

[IMAGE 209]
[IMAGE 210]
[IMAGE 211]
[IMAGE 212]

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(2)

According to question, roller is moved 50 cm i.e., $V_{center} \cdot t = 50 cm$

$Radius of roller, R = \frac{20}{2} = 10 cm and of axle, r = \frac{10}{2} = 5 cm$

For no slipping at the ground

$V_{center} = ω R$ $∴$ $Velocity of scale = (V_{center} + ω r)$

$∴$ $Distance moved by scale = (V_{center} + ω r) t$

$= (V_{center} + \frac{V_{center} r}{R}) t$ $(∵ ω = \frac{V}{R})$

$= \frac{3 V_{center}}{2} \cdot t = \frac{3}{2} \times 50 = 75 cm$

[IMAGE 213]

Q 3 :

A small object of uniform density rolls up a curved surface with an initial velocity $v$ . It reaches up to a maximum height of $\frac{3 v^{2}}{4 g}$ with respect to the initial position. The object is [2007]

[IMAGE 214]

ring
solid sphere
hollow sphere
disc

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(4)

Applying energy conservation principle of

$\frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2} = m g (\frac{3 v^{2}}{4 g})$

$∴$ $\frac{1}{2} m v^{2} + \frac{1}{2} I \frac{v^{2}}{R^{2}} = \frac{3}{4} m v^{2}$ $[∵ v = R ω]$

$∴$ $\frac{1}{2} I \frac{v^{2}}{R^{2}} = \frac{3}{4} m v^{2} - \frac{1}{2} m v^{2} = \frac{1}{4} m v^{2}$ $\Rightarrow I = \frac{1}{2} m R^{2}$

This is the moment of inertia of the disc hence the object is disc.

Q 4 :

A cylinder rolls up an inclined plane, reaches some height, and then rolls down (without slipping throughout these motions). The directions of the frictional force acting on the cylinder are [2002]

up the incline while ascending and down the incline while descending
up the incline while ascending as well as descending
down the incline while ascending and up the incline while descending
down the incline while ascending as well as descending

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(2)

[IMAGE 215]

Cylinder to be moving on a frictionless surface. In both the cases roll up roll down the acceleration of the centre of mass of the cylinder is $g \sin θ$ . Also no torque about the centre of cylinder is acting on the cylinder since we assumed the surface to be frictionless and the forces acting on the cylinder is mg and N which pass through the centre of cylinder. Therefore the net movement of the point of contact in both the cases is in the downward direction and hence the frictional force always acts in the upward direction.

Q 5 :

A cubical block of side $L$ rests on a rough horizontal surface with coefficient of friction $μ$ . A horizontal force $F$ is applied on the block as shown. If the coefficient of friction is sufficiently high so that the block does not slide before toppling, the minimum force required to topple the block is [2000]

[IMAGE 216]

infinitesimal
$\frac{m g}{4}$
$\frac{m g}{2}$
$m g (1 - μ)$

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(3)

[IMAGE 217]

Here, due to applied force normal reaction shifts to one corner. At this situation, the cubical block starts toppling about point O. About point O torque

$F \times L = m g \times \frac{L}{2} \Rightarrow F = \frac{m g}{2}$

Hence minimum force required to topple the block,

$F = \frac{m g}{2}$

Q 6 :

Two identical uniform discs roll without slipping on two different surfaces $A B$ and $C D$ (see figure) starting at $A$ and $C$ with linear speeds $v_{1}$ and $v_{2}$ , respectively, and always remain in contact with the surfaces. If they reach $B$ and $D$ with the same linear speed and $v_{1} = 3 m/s$ then $v_{2}$ in $m/s$ is ( $g = 10 {m/s}^{2}$ ) [2015]

[IMAGE 218]

(7)

$K . E_{total} of a rolling disc = K . E_{trans} + K . E_{rot}$

$= \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2} = \frac{1}{2} m v^{2} + \frac{1}{2} (\frac{1}{2} m R^{2}) (\frac{v^{2}}{R^{2}})$

$∴$ $K . E_{total} = \frac{3}{4} m v^{2}$

$For surface A B$

$K . E_{i} + loss in gravitational potential energy = K . E_{f}$

$\frac{3}{4} m {(3)}^{2} + m g (30) = \frac{3}{4} m V_{B}^{2} ...(i)$

$For surface C D$

$\frac{3}{4} m {(v_{2})}^{2} + m g (27) = \frac{3}{4} m V_{D}^{2} ...(ii)$

$∵$ $V_{B} = V_{D}$ $∴$ $from eq. (i) and (ii)$

$\frac{3}{4} m {(3)}^{2} + m g \times 30 = \frac{3}{4} m {(v_{2})}^{2} + m g \times 27$

$∴$ $v_{2} = 7 m/s$

Q 7 :

A boy is pushing a ring of mass $2 kg$ and radius $0.5 m$ with a stick as shown in the figure. The stick applies a force of 2N on the ring and rolls it without slipping with an acceleration of $0.3 {m/s}^{2}$ .

The coefficient of friction between the ground and the ring is large enough that rolling always occurs and the coefficient of friction between the stick and the ring is (P/10). The value of P is [2011]

[IMAGE 219]

(4)

The stick applies (2N) force so, point of contact O of the ring with ground tends to slide. But the frictional force $f_{2}$ does not allow this and creates a torque about 'C' which starts rolling the ring. Between the ring & the stick a friction force $f_{1}$ also acts.

[IMAGE 220]

$F - f_{2} = m a$

$2 - f_{2} = 2 \times 0.3 \Rightarrow f_{2} = 1.4 N$

$Applying τ = I α about C$

$(f_{2} - f_{1}) R = I α = I \frac{a}{R}$ $[∵ a = R α]$

$\Rightarrow [1.4 - μ \times 2] \times 0.5 = 2 \times {(0.5)}^{2} \times \frac{0.3}{0.5}$ $[∵ I = M R^{2}]$

$∴$ $μ = 0.4 = \frac{4}{10} = \frac{P}{10} \Rightarrow P = 4$

Q 8 :

At time $t = 0$ , a disk of radius $1 m$ starts to roll without slipping on a horizontal plane with an angular acceleration of $α = \frac{2}{3} {rad s}^{- 2}$ . A small stone is stuck to the disk. At $t = 0$ , it is at the contact point of the disk and the plane. Later, at time $t = \sqrt{π} s$ , the stone detaches itself and flies off tangentially from the disk. The maximum height (in m) reached by the stone measured from the plane is $\frac{1}{2} + \frac{x}{10}$ . The value of $x$ is _____. [Take $g = 10 {ms}^{- 2}$ ] [2022]

(0.52)

The angle rotated by disc in $t = \sqrt{π} sec is$

$θ = ω_{0} t + \frac{1}{2} α t^{2} = \frac{1}{2} \times \frac{2}{3} {(\sqrt{π})}^{2} = \frac{π}{3} rad$

Angular velocity at the end of $t = \sqrt{π} sec is$

$ω = ω_{0} + α t = \frac{2}{3} \sqrt{π} rad/s$

and, as it is pure rolling so $V_{c m} = ω R = \frac{2}{3} \sqrt{π} \times 1 m/s$

So, at the moment of detachment we get the following situation

[IMAGE 221]

So, $V = \sqrt{V_{c m}^{2} + {(ω R)}^{2} + 2 (ω R) (V_{c m}) \cos 120 °}$

$= V_{c m}$ $[∵ V_{c m} = ω R]$

$= \frac{2 \sqrt{π}}{3} m/s$

[IMAGE 222]

$\tan θ = \frac{(ω R) \sin 120 °}{V_{c m} + (ω R) \cos 120 °} = \frac{\frac{\sqrt{3}}{2}}{1 - \frac{1}{2}} = \sqrt{3}$

$∴$ $θ = 60 °$

$Then, H_{\max} = \frac{u^{2} \sin^{2} θ}{2 g} = \frac{{(\frac{2 \sqrt{π}}{3})}^{2} \times \frac{3}{4}}{2 \times 10} = \frac{\frac{4 π}{9} \times \frac{3}{4}}{20} = \frac{π}{60} m$

$Height from ground = R (1 - \cos \frac{π}{3}) + \frac{π}{60} = \frac{R}{2} + \frac{π}{60} = \frac{1}{2} + \frac{π}{60} = \frac{1}{2} + \frac{1}{10} (\frac{π}{6})$

$∴$ $x = \frac{π}{6} \approx 0.52$

Q 9 :

A solid sphere of mass 1 kg and radius 1 m rolls without slipping on a fixed inclined plane with an angle of inclination $θ = 30 °$ from the horizontal. Two forces of magnitude 1 N each, parallel to the incline, act on the sphere, both at distance $r = 0.5 m$ from the center of the sphere, as shown in the figure. The acceleration of the sphere down the plane is _______ ${ms}^{- 2}$ . (Take $g = 10 {ms}^{- 2}$ ) [2022]

[IMAGE 223]

(2.86)

[IMAGE 224]

We know that for a rolling body, point of contact is instantaneous axis of rotation.

$So, τ_{point of contact} = I α$

$\Rightarrow 1 \times 0.5 + (m g \sin 30 °) R - 1.5 \times 1 = I α$

$\Rightarrow 0.5 + 5 R - 1.5 = I α$

$\Rightarrow 5 R - 1 = \frac{7}{5} m R^{2} α$

$\Rightarrow 4 = \frac{7}{5} α$

$\Rightarrow α = \frac{20}{7} {rad/s}^{2}$

$So, a_{c m} = α R = \frac{20}{7} \times 1 = \frac{20}{7} {m/s}^{2} = 2.86 {m/s}^{2}$

Q 10 :

A ring and a disc are initially at rest, side by side, at the top of an inclined plane which makes an angle $60 °$ with the horizontal. They start to roll without slipping at the same instant of time along the shortest path. If the time difference between their reaching the ground is $\frac{2 - \sqrt{3}}{\sqrt{10}}$ s, then the height of the top of the inclined plane, in metres, is _______. Take $g = 10 {ms}^{- 2}$ . [2018]

(0.75)

The time taken to reach the ground or descend

$t = \frac{1}{\sin θ} \sqrt{\frac{2 h}{g} (1 + \frac{K^{2}}{R^{2}})}$

$For ring$ $t_{1} = \frac{1}{\sin 60 °} \sqrt{\frac{2 h}{g} (1 + 1)} = \frac{4}{\sqrt{3}} \sqrt{\frac{h}{g}}$ $∵$ $\frac{K^{2}}{R^{2}} = 1$

$For disc$ $t_{2} = \frac{1}{\sin 60 °} \sqrt{\frac{2 h}{g} (1 + \frac{1}{2})} = \sqrt{\frac{4 h}{g}}$ $∵$ $\frac{K^{2}}{R^{2}} = \frac{1}{2}$

$Given t_{1} - t_{2} = \frac{2 - \sqrt{3}}{\sqrt{10}}$

$∴$ $\frac{4}{\sqrt{3}} \sqrt{\frac{h}{g}} - \sqrt{\frac{4 h}{g}} = \frac{2 - \sqrt{3}}{\sqrt{10}}$ $\Rightarrow 2 \sqrt{h} - \sqrt{3} \sqrt{h} = \sqrt{3} - \frac{3}{2}$

$\Rightarrow \sqrt{h} (2 - 1.732) = 1.732 - 1.5 \Rightarrow \sqrt{h} = \frac{0.232}{0.268}$

$∴$ $h \approx 0.75 m (Height of the top of the inclined plane)$

Q 11 :

An annular disk of mass $M$ , inner radius $a$ and outer radius $b$ is placed on a horizontal surface with coefficient of friction $μ$ , as shown in the figure. At some time, an impulse $J_{0} \hat{x}$ is applied at a height $h$ above the center of the disk. If $h = h_{m}$ , then the disk rolls without slipping along the $x$ -axis. Which of the following statement(s) is(are) correct? [2023]

[IMAGE 225]

For $μ \neq 0$ and $a \to 0$ , $h_{m} = \frac{b}{2}$ .
For $μ \neq 0$ and $a \to 0$ , $h_{m} = b$ .
For $h = h_{m}$ the initial angular velocity does not depend on the inner radius $a$ .
For $μ = 0$ and $h = 0$ , the wheel always slides without rolling.

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Select one or more options

(1, 2, 3, 4)

Impulse $\vec{J} = Δ \vec{P}$ so

$J_{0} = m v ...(i)$

$J_{0} h_{m} = I_{c} ω \Rightarrow h_{m} = \frac{I_{c} ω}{J_{0}} = \frac{I_{c} ω}{m v} \Rightarrow h_{m} = \frac{I_{c}}{m R} [∵ V = R ω]$

[IMAGE 226]

$(1) If a = 0, I_{c} = \frac{1}{2} m b^{2} and R = b$ $∴$ $h_{m} = \frac{b}{2}$

$(2) If a = b, I_{c} = m b^{2} and R = b$ $∴$ $h_{m} = b$

$(3) v = \frac{J_{0}}{m} \Rightarrow 100 = \frac{V}{R} = \frac{J_{0}}{m R}$

$(4)$ As force is acting on centre of mass so no rotation for $μ = 0$ and $h = 0$ , the wheel always slides without rolling.

Q 12 :

A horizontal force $F$ is applied at the center of mass of a cylindrical object of mass $m$ and radius $R$ , perpendicular to its axis as shown in the figure. The coefficient of friction between the object and the ground is $μ$ . The center of mass of the object has an acceleration $a$ . The acceleration due to gravity is $g$ . Given that the object rolls without slipping, which of the following statement(s) is(are) correct? [2021]

[IMAGE 227]

For the same $F$ , the value of $a$ does not depend on whether the cylinder is solid or hollow
For a solid cylinder, the maximum possible value of $a$ is $2 μ g$
The magnitude of the frictional force on the object due to the ground is always $μ m g$
For a thin-walled hollow cylinder, $a = \frac{F}{2 m}$

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Select one or more options

(2, 4)

[IMAGE 228]

For solid cylinder,

$F \times R = \frac{3}{2} m R^{3} \times (\frac{a}{R}) \Rightarrow a = \frac{2 F}{3 m}$

For hollow cylinder,

$F \times R = (2 m R^{2}) \times \frac{a}{R} \Rightarrow a = \frac{F}{2 m}$

For solid cylinder

$f = F - m \times \frac{2 F}{3 m} = \frac{F}{3} \leq μ m g \Rightarrow F \leq 3 μ m g$

$∴$ $a_{\max} = 2 μ g$

Q 13 :

A wheel of radius $R$ and mass $M$ is placed at the bottom of a fixed step of height $R$ as shown in the figure. A constant force is continuously applied on the surface of the wheel so that it just climbs the step without slipping. Consider the torque $τ$ about an axis normal to the plane of the paper passing through the point $Q$ . Which of the following options is/are correct? [2017]

[IMAGE 229]

If the force is applied at point $P$ tangentially then $τ$ decreases continuously as the wheel climbs
If the force is applied normal to the circumference at point $X$ then $τ$ is constant
If the force is applied normal to the circumference at point $P$ then $τ$ is zero
If the force is applied tangentially at point $S$ then $τ \neq 0$ but the wheel never climbs the step

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(3)

[IMAGE 230]

If the force $(F)$ is applied at P tangentially then the torque remains constant and $τ = F \times 2 R$ .

If force is applied normal to X, then as the wheel climbs, then the perpendicular distance of force from Q will go on changing initially the perpendicular distance is QM, later it becomes QM'.

If the force (F) is applied normal to the circumference at point P then $τ = 0$ .

If the force (F) is applied tangentially at point S then $τ = F \times R$ and the wheel climbs.

Q 14 :

The figure shows a system consisting of (i) a ring of outer radius $3 R$ rolling clockwise without slipping on a horizontal surface with angular speed $ω$ and (ii) an inner disc of radius $2 R$ rotating anti-clockwise with angular speed $ω / 2$ . The ring and disc are separated by frictionless ball bearings. The point $P$ on the inner disc is at a distance $R$ from the origin, where $O P$ makes an angle of $30 °$ with the horizontal. Then with respect to the horizontal surface, [2012]

[IMAGE 231]

the point O has linear velocity $3 R ω \hat{i}$
the point P has linear velocity $\frac{11}{4} R ω \hat{i} + \frac{\sqrt{3}}{4} R ω \hat{k}$
the point P has linear velocity $\frac{13}{4} R ω \hat{i} - \frac{\sqrt{3}}{4} R ω \hat{k}$
the point P has linear velocity $(3 - \frac{\sqrt{3}}{4}) R ω \hat{i} + \frac{1}{4} R ω \hat{k}$

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Select one or more options

(1, 2)

Velocity at centre $' O'$ $∴$ ${\vec{v}}_{O} = 3 R ω \hat{i}$

[IMAGE 232]

${\vec{V}}_{P} = 3 R ω \hat{i} - \frac{R ω}{2} \sin 30 ° \hat{i} + \frac{R ω}{2} \cos 30 ° \hat{k}$

$∴$ ${\vec{V}}_{P} = [3 R_{ω} \hat{i} - \frac{R_{ω}}{4} \hat{i}] + \frac{\sqrt{3} R_{ω}}{4} \hat{k}$

$or, {\vec{V}}_{P} = \frac{11}{4} R_{ω} \hat{i} + \frac{\sqrt{3}}{4} R_{ω} \hat{k}$

Q 15 :

A sphere is rolling without slipping on a fixed horizontal plane surface. In the figure, A is the point of contact, B is the centre of the sphere and C is its topmost point. Then, [2009]

[IMAGE 233]

${\vec{V}}_{C} - {\vec{V}}_{A} = 2 ({\vec{V}}_{B} - {\vec{V}}_{C})$
${\vec{V}}_{C} - {\vec{V}}_{B} = {\vec{V}}_{B} - {\vec{V}}_{A}$
$| {\vec{V}}_{C} - {\vec{V}}_{A} |$ $= 2$ $| {\vec{V}}_{B} - {\vec{V}}_{C} |$
$| {\vec{V}}_{C} - {\vec{V}}_{A} |$ $= 4$ $| {\vec{V}}_{B} |$

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Select one or more options

(2, 3)

[IMAGE 234]

Here, ${\vec{V}}_{A} = 0$

${\vec{V}}_{B} = {\vec{V}}_{0}$

${\vec{V}}_{C} = 2 {\vec{V}}_{0}$

$∴$ ${\vec{V}}_{C} - {\vec{V}}_{B} = {\vec{V}}_{B} - {\vec{V}}_{A}$

$and$ $| {\vec{V}}_{C} - {\vec{V}}_{A} |$ $=$ $| {\vec{V}}_{B} - {\vec{V}}_{C} |$

Q 16 :

A solid cylinder is rolling down a rough inclined plane of inclination $θ$ . Then [2006]

The friction force is dissipative
The friction force is necessarily changing
The friction force will aid rotation but hinder translation
The friction force is reduced if $θ$ is reduced

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Select one or more options

(3, 4)

[IMAGE 235]

The sliding friction acts in the opposite direction of $m g \sin θ$ to oppose the relative motion. Because of frictional force the cylinder rolls.

Hence frictional force aids rotation but hinders translational motion.

Applying $F_{net} = m a$ along the direction of inclined plane, $m g \sin θ - f = m a_{c},$

where $a_{c}$ is acceleration of centre of mass of the cylinder

$∴$ $f = m g \sin θ - m a_{c} ...(i)$

$But a_{c} = \frac{g \sin θ}{1 + \frac{I_{c}}{m R^{2}}} = \frac{g \sin θ}{1 + \frac{m R^{2} / 2}{m R^{2}}} = \frac{2}{3} g \sin θ$

$∴$ $f = \frac{m g \sin θ}{3}$

Clearly, if $θ$ is reduced, frictional force is reduced.

Q 17 :

STATEMENT-1: Two cylinders, one hollow (metal) and the other solid (wood) with the same mass and identical dimensions are simultaneously allowed to roll without slipping down an inclined plane from the same height. The hollow cylinder will reach the bottom of the inclined plane first.

STATEMENT-2: By the principle of conservation of energy, the total kinetic energies of both the cylinders are identical when they reach the bottom of the inclined plane. [2008]

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
Statement-1 is True, Statement-2 is False
Statement-1 is False, Statement-2 is True

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(4)

In case of pure rolling on inclined plane acceleration of a body

$a = \frac{g \sin θ}{1 + \frac{I}{M R^{2}}}$

For hollow cylinder $\frac{I}{M R^{2}} = \frac{M R^{2}}{M R^{2}} = 1$

For solid cylinder $\frac{I}{M R^{2}} = \frac{\frac{1}{2} M R^{2}}{M R^{2}} = \frac{1}{2}$

Hence acceleration of solid cylinder is more than hollow cylinder and therefore solid cylinder will reach the bottom of the inclined plane first.

In the case of rolling there will be no work done by friction. Therefore total mechanical energy remains conserved.

$∴$ ${(K . E)}_{solid} = {(K . E)}_{hollow} = decrease in P.E = m g h$

Topic Question Set

Q 1 :

A disc is rolling without slipping with angular velocity $ω$ . P and Q are two points equidistant from the centre C. The order of magnitude of velocity is [2004]

[IMAGE 207]

Q 3 :

A small object of uniform density rolls up a curved surface with an initial velocity $v$ . It reaches up to a maximum height of $\frac{3 v^{2}}{4 g}$ with respect to the initial position. The object is [2007]

[IMAGE 214]

Q 4 :

A cylinder rolls up an inclined plane, reaches some height, and then rolls down (without slipping throughout these motions). The directions of the frictional force acting on the cylinder are [2002]

Q 15 :

A sphere is rolling without slipping on a fixed horizontal plane surface. In the figure, A is the point of contact, B is the centre of the sphere and C is its topmost point. Then, [2009]

[IMAGE 233]

Q 16 :

A solid cylinder is rolling down a rough inclined plane of inclination $θ$ . Then [2006]

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Topic Question Set

Q 1 : A disc is rolling without slipping with angular velocity ω. P and Q are two points equidistant from the centre C. The order of magnitude of velocity is [2004] [IMAGE 207]

Q 3 : A small object of uniform density rolls up a curved surface with an initial velocity v. It reaches up to a maximum height of 3v24g with respect to the initial position. The object is [2007] [IMAGE 214]

Q 4 : A cylinder rolls up an inclined plane, reaches some height, and then rolls down (without slipping throughout these motions). The directions of the frictional force acting on the cylinder are [2002]

Q 15 : A sphere is rolling without slipping on a fixed horizontal plane surface. In the figure, A is the point of contact, B is the centre of the sphere and C is its topmost point. Then, [2009] [IMAGE 233]

Q 16 : A solid cylinder is rolling down a rough inclined plane of inclination θ. Then [2006]

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Q 1 :

A disc is rolling without slipping with angular velocity $ω$ . P and Q are two points equidistant from the centre C. The order of magnitude of velocity is [2004]

[IMAGE 207]

Q 3 :

A small object of uniform density rolls up a curved surface with an initial velocity $v$ . It reaches up to a maximum height of $\frac{3 v^{2}}{4 g}$ with respect to the initial position. The object is [2007]

[IMAGE 214]

Q 4 :

A cylinder rolls up an inclined plane, reaches some height, and then rolls down (without slipping throughout these motions). The directions of the frictional force acting on the cylinder are [2002]

Q 15 :

A sphere is rolling without slipping on a fixed horizontal plane surface. In the figure, A is the point of contact, B is the centre of the sphere and C is its topmost point. Then, [2009]

[IMAGE 233]

Q 16 :

A solid cylinder is rolling down a rough inclined plane of inclination $θ$ . Then [2006]