Can any one solve this Physics 101 questions? 2. A horizontal cylinder of mass M

Question

Can any one solve this Physics 101 questions?

2. A horizontal cylinder of mass M and radius R on a frictionless has axle an unstretchable string wrapped around it. A mass m is attached to the other end of the string, and rests on a table below the cylinder, with the cylinder at rest and the string taut. The mass is then raised up a height h, the table is removed, and the mass is dropped. Just after the string becomes taut again... a) What is the speed of the mass m? (8 pts) b) What is the angular speed of the c (5 pts) c) Is this "collision" elastic? Prove your answer! (7 pts)

Explanation / Answer

(A) just before string becomes taut,

Applying energy conservation,

m g h = m v0^2 /2

v0 = sqrt(2 g h)

Applying angular momentum conservation for the system,

angular momentum just before string becomes taut = just after string becomes taut

m v0 R = m v R + I w

m v0 R = m v R + ( M R^2 / 2) (v / R)

m v0 R= m v R + M v R / 2

(2m + M) v R / 2 = m v0 R

v = 2 m sqrt(2 g h) / (2m + M)

(b) w = v / R

w = ( sqrt(2gh) / R) (2m / (2 m + M))

(C) Ki = m v^2 /2 = m g h

Kf = m v^2 /2 + I w^2 / 2

m v^2 / 2= m [2 m sqrt(2 g h) / (2m + M)]^2 /2

= 2 m^3 (2 g h) / (2m + M)^2

= m g h ( 2 m / (2m + M) )^2

I w^2 / 2 = ( M R^2 / 2) (( sqrt(2gh) / R) (2m / (2 m + M)) )^2 / 2

= ( M R^2 / 4) [ 8 m g h m^2 / (2m + M)^2]

= m g h [ 2 M m / (2m + M)]^2

Kf = m gh [ ( 2 m / (2m + M) )^2 + (sqrt(2 M m) / (2m + M))^2 ]

= m g h [ (4 m^2 + 2 M m) / (2m + M)^2]

m g h [ 2m (2m + M) / (2m + M)^2]

= m g h (2m / (2m + M))

2m / (2m + M) < 1

Kf < m gh

Kf < Ki

final energy is less then initial energy .

so there is energy lost (energy not conserved)

hence collision is not elastic.

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