A flywheel with a radius of 0.310 m starts from rest and accelerates with a cons

Question

A flywheel with a radius of 0.310m starts from rest and accelerates with a constant angular acceleration of 0.660rad/s2 .

Part A. Compute the magnitude of the tangential acceleration, the radial acceleration, and the resultant acceleration of a point on its rim at the start.

Part B. Compute the magnitude of the tangential acceleration, the radial acceleration, and the resultant acceleration of a point on its rim after it has turned through 60.0 degrees.

Part C. Compute the magnitude of the tangential acceleration, the radial acceleration, and the resultant acceleration of a point on its rim after it has turned through 120 degrees.

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A vacuum cleaner belt is looped over a shaft of radius 0.46cm and a wheel of radius 1.60cm . The arrangement of the belt, shaft, and wheel is similar to that of the chain and sprockets . The motor turns the shaft at rotational velocity ?= 58.0rev/s and the moving belt turns the wheel, that in turn is connected by another shaft to the roller that beats the dirt out of the rug being vacuumed. Assume that the belt doesn't slip on either the shaft or the wheel.

Part A. What is the angular velocity of the wheel in rad/s?

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A solid, uniform ball rolls without slipping up a hill. At the top of the hill, it is moving horizontally; then it goes over the vertical cliff. Take V = 27.0m/s and H = 25.0m .

Part A. How fast is it moving just before it lands?

Part B. Notice that when the ball lands, it has a larger translational speed than it had at the bottom of the hill. Does this mean that the ball somehow gained energy by going up the hill? Explain!

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Explanation / Answer

The formulae for components of acceleration of a particle moving in a curve with polar co-ordinates (r, t) are:
Normal: r''- r(t')^2
Tangential: rt''+ 2r't'

For circular motion r' and r'' are zero, yielding:
Normal: - r(t')^2
Tangential: rt''

For the flywheel, the normal acceleration of a particle on the rim at the start is:
0, as t' is initially 0.

The tangential acceleration is:
0.300 * 0.600
= 0.18 m/s^2.

The resultant acceleration is the same as the tangential.

When the flywheel has rotated through pi/3 (= 60 deg):
(t')^2 = 2 * 0.600 * pi/3
t = 1.12 rad/s.

The normal acceleration is:
- 0.300 * 2 * 0.6 * pi / 3
= - 0.38 m/s^2.

The tangential acceleration is still 0.18 m/s^2, as it is constant.

The resultant accleration is of magnitude:
sqrt(0.38^2 + 0.18^2)
= 0.42 m/s^2.

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