Answered

A cylinder of mass m and radius R has a momentvof inertia of 1/2mr squared. The cylinder is released from rest at a height h on an without slipping. What is the velocity of the cylinder when it reaches the bottom of the incline?

Sagot :

Answer:

SOLUTION ONLY!

Explanation:

Calculate the angular velocity of the cylinder when it reaches the bottom of the inclined plane.

Take H = 7 m, M = 0.25 kg, R = 0.5 m, and q = 30°. Do NOT use Newton's laws of motion.

Conservation of energy yields KE(translation)+ KE (rotation) = PE e(it top of inclined plane)

(1)

1

2

MV2 + 1

2

Icylinder w2 = M g H

Using the connection between angular velocity w and translational velocity V=wR or w = V/R is used together with the

moment of ineretial of a cylinder Icylinder=(1/2)MR2 yields

(2)

1

2

MV2 + 1

2 IH1 ê 2L M R2M HV ê RL2 = M g H

(3)

1

2

MV2 + 1

4

MV2 = M g H and thus

3

4

V2 = g H and finally V = 4

3

gH

Substitution of the numbers for g=9.8 m/s2 and H=7m yields for the linear velocity V= 9.6 m/s .

V = 4

3 * 9.8 * 7

9.56382

and for the angular velocity w=4.8 Rad/sec.

w = 9.6 ê 0.5

19.2

Viz Other Questions