Question

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?

Answer #1

The system is like annular disc. Moment of inertia of this
system is I = ½ * M(r_{1}^{2} +
r_{2}^{2})

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

I = ½ * 1 * (0.080^{2} + 0.060^{2}) kg
m^{2}

I = 0.005 kg m^{2}

Total Kinetic Energy is KE

KE = ½mv² + ½I?² v = r_{1} ?

= ½mv² + ½I(v/r_{1})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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