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Vlada [557]
3 years ago
6

A person is riding on a Ferris wheel of radius R. He starts at the lowest point of the wheel. When the wheel makes one complete

revolution, is the net work done by the gravitational force positive, zero or negative? Do you need to know how the speed of the person changed before you can answer the question?
Physics
1 answer:
Wittaler [7]3 years ago
4 0

Answering the two questions in reverse order:

-- No. I don't need to know how the speed of the person changed before I can answer the question.  I can answer it now.

-- The NET work done by the gravitational force is<em> zero</em>.

-- As the person and his girl-friend go up the first half of the wheel, the motor does positive work and gravity does negative work.

-- After they pass the peak at the top and come down the second half of the wheel, the motor does negative work and gravity does positive work, even though the couple may be interested in other things during that time.

-- The total work done by gravity in one complete revolution is zero.

-- The total work done by the motor in one complete revolution is only what it takes to pay back the energy robbed by friction and air resistance.

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Yes

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3 years ago
The 480 g bar is rotating as shown what is the angular momentum of the bar about the axle?
Greeley [361]
On a similar problem wherein instead of 480 g, a 650 gram of bar is used:

Angular momentum L = Iω, where 
<span>I = the moment of inertia about the axis of rotation, which for a long thin uniform rod rotating about its center as depicted in the diagram would be 1/12mℓ², where m is the mass of the rod and ℓ is its length. The mass of this particular rod is not given but the length of 2 meters is. The moment of inertia is therefore </span>
<span>I = 1/12m*2² = 1/3m kg*m² </span>

<span>The angular momentum ω = 2πf, where f is the frequency of rotation. If the angular momentum is to be in SI units, this frequency must be in revolutions per second. 120 rpm is 2 rev/s, so </span>
<span>ω = 2π * 2 rev/s = 4π s^(-1) </span>

<span>The angular momentum would therefore be </span>
<span>L = Iω </span>
<span>= 1/3m * 4π </span>
<span>= 4/3πm kg*m²/s, where m is the rod's mass in kg. </span>

<span>The direction of the angular momentum vector - pseudovector, actually - would be straight out of the diagram toward the viewer. </span>

<span>Edit: 650 g = 0.650 kg, so </span>
<span>L = 4/3π(0.650) kg*m²/s </span>
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