(a) The moment of inertia of a rolling marble is I = MR 2 , where M is the mass

Question

(a) The moment of inertia of a rolling marble is

I =

MR2,

where M is the mass of the marble and R is the radius. The marble is placed in front of a spring that has a constant k and has been compressed a distance

xc.

The spring is released and as the marble comes off the spring it begins to roll without slipping. Note: The static friction that causes rolling without slipping does not do work. Derive an expression for the time it takes for the marble to travel a distance D along the surface after it has lost contact with the spring. (Use the following as necessary:

xc, M, D, and k.)

t =

Explanation / Answer

Let v is the speed of the marble when it leaves the contact with the spring.

Apply conservation of enrgy

initial elastic potential energy = final kinetic enrgy of the marble

0.5*k*xc^2 = 0.5*M*v^2 + 0.5*I*w^2

0.5*k*xc^2 = 0.5*M*v^2 + 0.5*((2/5)*M*R^2)*w^2

0.5*k*xc^2 = 0.5*M*v^2 + 0.2*M*(R*w)^2

0.5*k*xc^2 = 0.5*M*v^2 + 0.2*M*v^2

0.5*k*xc^2 = 0.7*M*v^2

v = sqrt(0.5*k*Xc^2/(0.7*M) )

time taken, t = D/v

= D/sqrt(0.5*k*Xc^2/(0.7*M) )

= D*sqrt(0.7*M/(0.5*k*Xc^2)) <<<<<---Answer

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