A plank of mass M is placed over a smooth inc | vedclass

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(D) Let the friction force acting on the sphere be $f$ directed up the plane.For the plank of mass $M$,the equation of motion along the plane is:$M a_

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zero

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(D) Let the friction force acting on the sphere be $f$ directed up the plane.For the plank of mass $M$,the equation of motion along the plane is:$M a_1 = M g \sin 	heta + f$  --- $(1)$where $a_1$ is the acceleration of the plank down the plane.For the sphere of mass $m$,let $a_2$ be its acceleration relative to the plank down the plane. The torque equation about the center of the sphere is:$f \cdot r = I \alpha = (\frac{1}{2} m r^2) \alpha$Since the sphere rolls without slipping on the plank,$a_2 = \alpha r$,so $f = \frac{1}{2} m a_2$,which gives $a_2 = \frac{2f}{m}$.The equation of motion for the sphere relative to the plank is:$m a_2 = m g \sin 	heta - f - m a_1$  --- $(2)$Substituting $a_1 = g \sin 	heta + \frac{f}{M}$ from $(1)$ into $(2)$:$m (\frac{2f}{m}) = m g \sin 	heta - f - m (g \sin 	heta + \frac{f}{M})$$2f = m g \sin 	heta - f - m g \sin 	heta - \frac{m f}{M}$$3f = - \frac{m f}{M}$$f (3 + \frac{m}{M}) = 0$Since $(3 + \frac{m}{M}) 
eq 0$,we must have $f = 0$.

Quantities	Their numerical values (in $SI$ units)
a. Angular acceleration of smaller wheel	$1$. $5/3$
b. Torque on the larger wheel	$2$. $100/3$
c. Angular acceleration of larger wheel	$3$. $5/2$

$A$ plank of mass $M$ is placed over a smooth inclined plane and a sphere of mass $m$ is also placed over the plank. Friction is sufficient between the sphere and the plank. If the plank and the sphere are released from rest,the frictional force on the sphere is:

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