Question

A Yo-Yo of mass $m$ has an axle of radius $b$ and a spool of radius $R$. The Yo-Yo is placed upright on a table and string is pulled with a horizontal force $\overrightarrow{F}$ to the right as shown in the figure.
The static friction coefficient between the Yo-Yo and the table is ${\mu }_s$. Now there are two cases
Case (a): The string is pulled very gently.
Case (b): The string is pulled with a jerk.
The Yo- Yo will rotate

• A. clockwise in case (a)
• B. anticlockwise in case (a)
• C. clockwise in case (b)
• D. anticlockwise in case (b)

Answer 1

Answer: (A), (D)

anticlockwise in case (b)

In first case, friction will act at the point of contact. Therefore, we will find the net torque about instantaneous axis of rotation (i.e throught point of contant). Due to this we don't have to consider torque due to friction in this case.

$\tau =I\alpha$

Case (a)
$F(R-b)=I_{\text{inst}}{\alpha }_1(\text{clockwise torque)}\Rightarrow {\alpha }_1=-F(R-b)/I_{\text{inst}}\text{(clockwise acceleration)}$

In the second case, due to sudden jerk, the force $F$ will become impulsive force due to which the yo-yo will attain an angular velocity. Friction force in this case is non-impulsive so it will not contribute for the rotation of the body. Hence, we can take the net torque about CoM itself.

Case (b)
$F^{}b=I_{\text{centre}}{\alpha }_2(\text{anticlockwise torque)}\Rightarrow {\alpha }_2=F^{\ast }b/I_{\text{centre}}$(anticlockwise acceleration)