A thin light string is wrapped around the outer rim of a uniform hollow cylinder

Question

A thin light string is wrapped around the outer rim of a uniform hollow cylinder of mass 4.00 kg having inner and outer radii as shown in the figure (Figure 1) . The cylinder is then released from rest.

A- How far must the cylinder fall before its center is moving at 6.91 m/s ?

d= m

B-If you just dropped this cylinder without any string, how fast would its center be moving when it had fallen the distance in part A?

v= m/s

c- Why do you get two different answers? The cylinder falls the same distance in both cases.

Essay answers are limited to about 500 words (3800 characters maximum, including spaces)

Explanation / Answer

Part A:

m = 4 kg , v =6.91 m/s

r1 =20 cm , r2 = 35 cm

d = (1+(1/2)(1+(r1/r2)^2)v^2/2*9.8

d = ((1+(1/2)(1+(20/35)^2))6.91^2/(2*9.8)

d =4.052 m

Part B:

v = (1+((1/2)(1+(r1/r2)^2))^0.5*V

v = 6.91*(1+(1/2)(1+(20/35)^2))^0.5

V = 8.9 m/s

Part C: remains same

In part (a) the cylinder has rotational as well as traslational kinetic energy and therefore less translational speed at a given kinetic eenergy. The kinetic energy comes from a decrease in gravitational potential energy and that is the same so in (a) the translational

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