A solid, uniform cylinder with mass M = 0.8 kg and radius R = 3 cm is released f

Question

A solid, uniform cylinder with mass M = 0.8 kg and radius R = 3 cm is released from the top of an inclined of height h = 0.6 m length L = 1.9 m.

A) What is the speed of the center of mass of the cylinder as it reaches the bottom of the incline, in m/s?

B)If the mass of the cylinder were increased by a factor of 3, the speed at the bottom of the incline would change by a factor of

C)If the radius of the cylinder were increased by a factor of 6, the speed at the bottom of the incline would change by a factor of

D)f the length of the incline were increased by a factor of 5 (while keeping the same height but changing the inclination), the speed of the cylinder at the bottom of the incline would change by a factor of

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Explanation / Answer

A) Moment of Inertia at the point of rolling = 2 MR^2

M = 0.8 kg R = 3 cm = 3 * 10^-2 m

Moment of inertia = I = 2 * 0.8 * 3 *10^-2 = 4.8 * 10^-2

Now loss in potential energy = gain in total kinetic energy = Mg h = 0.6 * 0.8 * 9.8 = 0.5 * I * w^2 ..... (1)

Solving above equation we get w = 14 rad/s

As there is no slippping speed = w*r = 14 * 3 * 10^-2 = 0.42 m/sec

B) According to the equation (1)

Mgh = 0.5 * 2 M * R ^2 * w ^2 .....(2)

as M gets cancelled out

M doesnt effect the speed

There fore speed doesnt change

C) From equation (2) keeping everythng else constant w is inversely prportional to R

Therefore if radius is increased by 6 the speed would decrease by factor of six i.e., becomes 1/6th of the original

D) As we can see from the equation (2) the speed doesnt depend on the lenght of the incline .

Therefore speed dopesn change

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