Question

A 1.0-kg ring with an inner radius of 0.060 m and an outer radius of 0.080...

A 1.0-kg ring with an inner radius of 0.060 m and an outer radius of 0.080 m is sent rolling without slipping up a ramp that makes an angle of 30? with the horizontal. The initial speed of the ring is 3.8 m/s

How far along the ramp does it travel before it comes to a stop?

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Answer #1

The system is like annular disc. Moment of inertia of this system is I = ½ * M(r12 + r22)

r1 = outer radius r2 = inner radius , M mass , velocity = 3.8 m/s

I = ½ * 1 * (0.0802 + 0.0602) kg m2

I = 0.005 kg m2

Total Kinetic Energy is KE

KE = ½mv² + ½I?² v = r1 ?

= ½mv² + ½I(v/r1)²

=  ½ * 1 * 3.8 * 3.8 +  ½ * 0.005 * (3.8 / 0.08)2

= 7.22 +5.64

= 12.86 J

All it converts to Potential energy

PE = m g h

12.86 = m g h

h = 12.86 / m g

= 1.312 m

How far along the ramp does it travel before it comes to a stop?

Since it is a ramp

d = 1.312 / sin 30

d = 2.624 m

It is 2.624 m along the ramp it travel before it comes to a stop.

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