The moment of inertia of a round object about an axis through its center is give

Question

The moment of inertia of a round object about an axis through its center is given by I = 1/3MR^2. The round object has mass M = 5.4 kg and radius R = 0.46 m, and rolls without slipping on a horizontal surface with an initial angular velocity of w0(omega not) = 4.783 rad/s. 1) What is the total Ke of the round object as it rolls on the hroizontal surface? 2) What fraction of the total Ke of the round object is due to rotation? 3) Now suppose the round object rolls up an inclined plane and then rolls back down again. The angle of the incline with respect to horizontal is theta = 33°, the round object does not slip on the incline. As the round object rolls down the ramp, what is the magnitude of its acceleration?

Explanation / Answer

here,

mass , m = 5.4 kg

R = 0.46 m

w0 = 4.783 rad/s

1)

the total kinetic energy of the object , TE = 0.5 * m*v^2 + 0.5 * I*w0^2

TE = 0.5 * m *( r*w)^2 + 0.5 * ( mR^2 /3) * w0^2

TE = 0.5 * 5.4*(0.46*4.783)^2 + 0.5 * ( 5.4 * 0.46^2 /3) * 4.783^2

TE = 17.43 J

2)

the fraction of toatl kinetic energy is the rotation energy,

f = KEr /TE

f= (0.5 * ( mR^2 /3) * w0^2) / 17.43

f = (0.5 * ( 5.4 * 0.46^2 /3) * 4.783^2) /17.43

f = 1/4

the fraction of toatl kinetic energy is the rotation energy is 1:4 or 0.25

3)

let the height reached be h

using conservation of energy

17.43 = 5.4 * 9.8*h

h = 0.33 m

distance travelled on incline , d = 0.33 / sin(33)

d = 0.605 m

v0 = r*w0

v0 = 2.2 m/s

let the accelration be a

using thrid equation o0f motion

v^2 - v0^2 = 2*d*a

0 - 2.2^2 = 2*0.605 * a

a = 4 m/s^2

the magnitude of the accelration is 4 m/s^2

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