JEE Advanced Physics Circular Motion & Banking of Road PYQ

Q 1 :

A ball of mass (m) 0.5 kg is attached to the end of a string having length (L) 0.5 m. The ball is rotated on a horizontal circular path about a vertical axis. The maximum tension that the string can bear is 324 N. The maximum possible value of angular velocity of the ball (in radian/s) is [2011]

9
18
27
36

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(4)

$T \sin θ = m R ω^{2}$ ...(i)

$T \cos θ = m g$ ...(ii)

Dividing (ii) by (i), we get

$\tan θ = \frac{ω^{2}}{R g} \Rightarrow ω = \sqrt{R g \tan θ}$

Clearly, $ω$ is maximum when $\tan θ$ is maximum, i.e., $θ = 90 °$

So, $T \sin 90 ° = m R ω^{2}$

$T = m L ω^{2} [Here, R = L]$

$\Rightarrow ω = \sqrt{\frac{T}{m L}} = \sqrt{\frac{324}{0.5 \times 0.5}}$

$= \frac{18}{0.5} = 36 rad/s$

Q 2 :

A small block is shot into each of the four tracks as shown below. Each of the tracks rises to the same height. The speed with which the block enters the track is the same in all cases. At the highest point of the track, the normal reaction is maximum in [2001]

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(1)

According to question, the speed with which the block enters the track is the same in all the tracks and the block rises to the same height, so from the law of conservation of energy, speed of the block at the highest point will be same in all four cases.

Let the velocity at the highest point be $v$ .

$(N + m g) provides the centripetal force \frac{m v^{2}}{R} to the body$

$N + m g = \frac{m v^{2}}{R}$

$or N = \frac{m v^{2}}{R} - m g$

$R$ (the radius of curvature) in the first case is minimum.

Hence, normal reaction N will be maximum in the first case.

Q 3 :

A thin circular coin of mass 5 gm and radius $\frac{4}{3} cm$ is initially in a horizontal $x y$ -plane. The coin is tossed vertically up ( $+ z$ direction) by applying an impulse of $\sqrt{\frac{π}{2}} \times 10^{- 2}$ N - s at a distance $\frac{2}{3} cm$ from its center. The coin spins about its diameter and moves along the $+ z$ direction. By the time the coin reaches back to its initial position, it completes $n$ rotations. The value of $n$ is ________.

[Given: The acceleration due to gravity $g = 10 {ms}^{- 2}$ ] [2023]

(30)

From impulse--momentum theorem,

$J = M V_{C M} \Rightarrow V = \frac{J}{M} = \frac{\sqrt{π / 2}}{100 \times \frac{5}{1000}} = \sqrt{2 π} m/s$

Total time taken, $t = \frac{2 v}{g}$

$= \frac{2 \times \sqrt{2 π}}{g} = \frac{2 \times \sqrt{2 π}}{10} = \frac{\sqrt{2 π}}{5} s$

By angular impulse--momentum theorem,

$J \times \frac{R}{2} = I_{C} ω = [\frac{1}{4} M R^{2}] ω$

$∴$ $ω = \frac{J \times \frac{R}{2}}{\frac{M R^{2}}{4}} = \frac{J \times 2}{M R}$

$= \frac{\frac{\sqrt{π / 2}}{100} \times 2}{\frac{5}{1000} \times \frac{4}{3} \times \frac{1}{100}} = 2 \times 75 \sqrt{2 π} rad/s$

$∵$ $θ = 2 π n = ω t \Rightarrow n = \frac{ω t}{2 π}$

$∴$ $n = \frac{2 \times 75 \sqrt{2 π} \times \frac{\sqrt{2 π}}{5}}{2 π} = 30$

Q 4 :

A wire, which passes through the hole in a small bead, is bent in the form of a quarter of a circle. The wire is fixed vertically on ground as shown in the figure. The bead is released from near the top of the wire and it slides along the wire without friction. As the bead moves from A to B, the force it applies on the wire is [2014]

always radially outwards
always radially inwards
radially outwards initially and radially inwards later
radially inwards initially and radially outwards later

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(4)

Suppose 'N' is acting radially outward.

Then, $m g \cos θ - N = \frac{m v^{2}}{R}$

$\Rightarrow N = m g \cos θ - \frac{m v^{2}}{R}$ ...(i)

And by energy conservation,

$\frac{1}{2} m v^{2} = m g [R - R \cos θ]$

$∴$ $\frac{v^{2}}{R} = 2 g (1 - \cos θ)$

Putting this value of $\frac{v^{2}}{R}$ in eqn. (i),

$N = m g \cos θ - m [2 g - 2 g \cos θ]$

$\Rightarrow N = m g \cos θ - 2 m g + 2 m g \cos θ$

$\Rightarrow N = 3 m g \cos θ - 2 m g \Rightarrow N = m g (3 \cos θ - 2)$

Clearly when $\cos θ > \frac{2}{3}$ , N is positive and acts radially outward.

So, force on wire is inward and if $\cos θ < \frac{2}{3}$ , N acts radially inward.

So, force on wire is outward.

Topic Question Set

Q 2 :

A small block is shot into each of the four tracks as shown below. Each of the tracks rises to the same height. The speed with which the block enters the track is the same in all cases. At the highest point of the track, the normal reaction is maximum in [2001]

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Topic Question Set

Q 2 : A small block is shot into each of the four tracks as shown below. Each of the tracks rises to the same height. The speed with which the block enters the track is the same in all cases. At the highest point of the track, the normal reaction is maximum in [2001]

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Q 2 :

A small block is shot into each of the four tracks as shown below. Each of the tracks rises to the same height. The speed with which the block enters the track is the same in all cases. At the highest point of the track, the normal reaction is maximum in [2001]