Question

please help with problem 53

Rotational Motion Problem Solving "An expert is a person who has made all the mistakes that can be made in a very narrow field." -Niels Bohr 53. An inclined plane makes an angle of with the horizontal, as shown above. A solid sphere of radius R and mass M is initially at rest in the position shown, such that the lowest point of the sphere is a vertical height h above the base of the plane. The sphere is released and rolls down the plane without slipping. The moment of inertia of the sphere about an axis through its center is 2MR 5. Express your answers in terms of M, R. h, g, and e. a. Draw a freebody diagram on the dot below of all the forces acting on the sphere from their point of origin. VO b. Determine the following for the sphere when it is on the inclined plane. i. Its linear acceleration (nole use both rotational and translational force equations) ii. The magnitude of the frictional force acting on it c. The solid sphere is replaced by a hollow sphere of identical radius R and mass M. The hollow sphere, which is released from the same location as the solid sphere, rolls down the incline without slipping. State whether the time for the hollow sphere is to reach the bottom of the inclined plane is greater than, less than or equal to that of the solid sphere. Justify your answer.

Answer 1

a. I hope you can easily draw the freebody diagram

There are mainly 3 forces acting on the body

1. Frictional force acting parallel to the plane and in the opposite direction of motion

2. Normal force which is perpendicular to the plane

3. Weight of the body that is gravitational force

Which can be resolved into two a cos$\Theta$ component acting in opposite direction of normal force and cancels it

And a sin$\Theta$ component opposite to frictional force

B. Here torque = f × r ( f = frictional force' r = radius of sphere ) = force × perpendicular distance

We have $\tau$ = Io × $\alpha$ where Io = moment of inertia and $\alpha$ = angular acceleration

$\tau$= (2MR^2 /5) × a/R = 2MRa /5

We could find a from this equation but $\tau$ is not given so we substitute $\tau$ by f R we get  f = 2Ma/5 equation 1

we can calculate f from equation of translational motion

According to newton's law  ma = f - mgsin$\Theta$ ( f= $\mu$ × normal force = $\mu$ mgcos$\theta$)

  f = m(a+gsin$\Theta$) equation 2

from equation 1 and 2

2Ma/5 = Ma+Mgsin$\Theta$

a = -5gsin$\Theta$/3

Putting these result into equation 2

We get f= -2gsin$\Theta$/3

C. hollow sphere has larger moment of inertia than a solid sphere it will have smaller acceleration and will take more time than solid sphere to reach gtounf