A solid cylinder with a mass of 4.5 kg, a radius of 0.017m, an initial velocity

Question

A solid cylinder with a mass of 4.5 kg, a radius of 0.017m, an initial velocity of 0.0 m/s, an initial angular velocity of 0.0 rad/s, and an initial height of 3.6 m, rolls down an incline until it hits the bottom where its height is 0.0m.

1. What is the cylinder's moment of inertia?

2. What is the cylinder's initial translational kinetic energy?

3. What is the cylinder's initial rotational kinetic energy?

4. What is the cylinder's initial gravitational potential energy?

5. What is the cylinder's final gravitational energy?

6. What is the cylinder's final velocity?

7. What is the cylinder's final angular velocity?

8. What is the cylinder's final translational kinetic energy?

9. What is the cylinder's final rotational kinetic energy?

10. What is the cylinder's total final mechanical energy?

Explanation / Answer

1) moment of interia , I = (1/2)*m*r2

I = (1/2)*4.5*0.0172 = 6.50*10-4 kg-m2

2) initial translation K.E ,   KT = (1/2)*m*Vi2

                                   = (1/2)*m*0 = 0 J

3) initial rotational K.E,    KR = (1/2)*I*wi2

                                       = 0                                ( wi ( initial angular velocity) = 0)

4) initial gravitational P.E = mgh

P.Ei = 4.5*9.8*3.6 = 158.76 J

5) final gravitational P.E = mg*0

                     = 0                              ( h= 0 since it is at bottom)

10) total mechanical energy = K.E + P.E

at top ie at height h it has only P.E and K.E = 0

at top . T.E = 0 + 158.76 = 158.76 J

at bottom , T.E = K.E + P.E                          ( here P.E = 0 because h=0)

at bottom , T.E = K.E

T.E = KT + KR

since energy remain conserved , so energy at top = energy at bottom = 158.76 J

158.76 = (1/2)*m*Vf2 + (1/2)*I*wf2

158.76 = (1/2)*m*Vf2 + (1/2)*I*m* ( Vf / r)2                                      ( v= wr)

on solving this , Vf ( final translation energy) = 6.85 m/s                 ------------------ans-6

7) wf ( final angular velocity) = Vf / r

                   = 6.85/ 0.017 = 402.9 rad/sec

8) final translation K.E , KT = (1/2)*m*Vf2

                      = (1/2)*4.5*6.852 = 105.57 J

9) final rotational K.E,   KR = (1/2)*I*wf2

                   = (1/2)*6.50*10-4*402.92

                    = 52.75 J

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