Question

A wheel $A$ starts rolling up a rough inclined plane another identical wheel $B$ starts rolling up a smooth plane having same inclination with the horizontal. If initial velocity of both the wheels is same then:

• A. $A$ stops ascending earlier than $B$
• B. kinetic energy of $B$ never becomes zero
• C. maximum height ascended by $A$ is less than that by $B$
• D. friction acting on $A$ remains constant during the round trip

Answer 1

Answer: (B), (D)

kinetic energy of $B$ never becomes zero
friction acting on $A$ remains constant during the round trip

Let $m$ and $h$ be the mass and maximum height respectively ascended by the sphere. Initial velocity of both the sphere is $v$ and angular velocity be $w$.

$v=Rw$ (pure rolling)

As $v$ decreases due to retardation,, thus $w$ will have to decrease to ensure pure rolling. And hence to do so friction force must act in upward direction. Similarly for the round trip $v$ will increase due to acceleration and thus $w$ will also increase (pure rolling) and hence friction force will act in upward direction.

Work-energy theorem: $W_g=\Delta K.E$

For sphere A: At maximum height, linear and angular velocity will be zero.

$-m_{A}g{h}_A=0-\frac{1}{2}I{w}^2-\frac{1}{2}{m}_A{v}^2$

$-m_{A}g{h}_A=0-\frac{1}{2}\times \frac{2}{5}{m}_A{R}^2{w}^2-\frac{1}{2}{m}_A{v}^2$

$m_{A}g{h}_A=\frac{1}{5}{m}_A{v}^2+\frac{1}{2}{m}_A{v}^2$

$⟹g{h}_A=\frac{7}{10}{v}^2$

For sphere B: At maximum height, only linear velocity will be zero and angular velocity will remain same as initial because there is no friction to change it.Thus B will have rotational K.E at maximum height and hence K.E of B can never be zero.

$-m_{B}g{h}_B=\frac{1}{2}I{w}^2-\frac{1}{2}I{w}^2-\frac{1}{2}{m}_A{v}^2$

$m_{B}g{h}_B=\frac{1}{2}{m}_B{v}^2$

$⟹g{h}_B=\frac{1}{2}{v}^2$

Thus $h_A>h_B$