(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.)

(b) Show that your answer for part (a) has the correct units.

Explanation / Answer

PE of the spring marble system = 0.5*K*xc2

after losing the contact KE of the marble = 0.5*M*v^2 + 0.5*I*w2

when a marble is rolling v = R*w , I = (2/5)*M*R2

KE = 0.5*M*v2 + 0.5*(2/5)*M* R 2* W2

KE = 0.5*M*v 2 + (1/5)*M*v2

KE = 0.7*M*V2

from energy conservation

KE = PE

0.7*M*v2 = 0.5*k*xc2

v = sqrt(5*K/7M)*xc

t = D/v

t = D/sqrt(5*K/7M)*xc

--------------------

units of D = m

units of k = N/m = kg s-2

units of M = kg

unit of xc = m

unit of time = s

7 & 5 are numbers, so they have no units

v = sqrt(kg*s-2/kg)*m = m s-1

t = D/v = m /ms-1 = s

the epression has correct units

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