A wheel of a bullock cart is rolling on a level road as shown in the figure below. If its linear speed is in the direction shown, which one of the following options is correct (P and Q are any highest and lowest points on the wheel respectively)?

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
A wheel of a bullock cart is rolling on a level road as shown in the figure below. If its linear speed is $v$ in the direction shown, which one of the following options is correct (P and Q are any highest and lowest points on the wheel respectively) [2024]

1 Point P moves slower than point Q.

2 Point P moves faster than point Q.

3 Both the points P and Q move with equal speed.

4 Point P has zero speed.

Ans.
(2)

For pure rolling $v_{c m} = r ω$

Given, $v_{c m} = v$

$∴$ $v = r ω$ ...(i)

Net velocity is the vector sum of translational and rotational velocity and $θ$ is the angle between translational and rotational velocity.

$| {\vec{v}}_{net} |$ $= \sqrt{v^{2} + {(r ω)}^{2} + 2 (v) (r ω) \cos θ}$ ...(i)

For point Q, $θ = 180 °$

$| {\vec{v}}_{Q} |$ $= \sqrt{v^{2} + r^{2} ω^{2} + 2 r v ω \cos 180 °}$

$= \sqrt{{(v - r ω)}^{2}} = v - r ω$

$| {\vec{v}}_{Q} |$ $= r ω - r ω = 0 (using (i))$ ...(ii)

Now for point P, $θ = 0 °$

Using eq. (i), we have

$| {\vec{v}}_{P} |$ $= \sqrt{v^{2} + {(r ω)}^{2} + 2 (v) (r ω) \cos 0 °}$

$= \sqrt{{(v + r ω)}^{2}} = v + r ω$

Using eq. (i), we have

$| {\vec{v}}_{P} |$ $= 2 r ω$ ...(iii)

So, it is clear from equation (ii) and (iii) that point P moves faster than point Q.

Want Us to Email you About Special Offers & Updates?