[eskimojoe2200]

I am completely lost on 4 of my homework problems. I just need to know how to set them up. We went through this chapter so fast that it's hard for me to remember exactly what to do.

1. A uniform 13-kg trap door is oriented horizontally and hinged as shown. (It has a length of 0.75 m) What is the magnitude of the torque on the door at the instant that the release is activated and the door can freely rotate?

2. Three forces are applied to a wheel of radius 0.350 m. One force is perpendicular to the rim, one is tangent to it, and the other one makes a 40 degree angle with the radius. What is the net torque on the wheel due to these three forces for an axis perpendicular to the wheel and passing through it's center? (The force perpendicular to the rim = 11.9 N; The force making a 40 degree angle = 14.6N; The force that is tanget to it = 8.50 N)

3. A playground mery-go-round has radius 2.40 m and moment of inertia 2100 kg. m2 about a vertical axis through it's center, and it turns with negligible friction. A child applies an 18.0 N force tangetially to the edge of the merry-go-round for 15 s. If the merry-go-round is initially at rest:
a. what is its angular speed after this 15 s interval
b. how much work did the child do on the merry-go-round
c. What is the average power supplied by the child.

4. The outstretched hands and arms of a figure skater preparing for a spin can be considered as a slender rod pivoting about an axis through its center. When the skater's hands and arms are brought in and wrapped around his body to execute the spin, the hands and arms can be considered as a thin-walled, hollow cylinder. His hands and arms have a combined mass 8.0 kg. When outstretched theys pan 1.8 m; when wrapped, they form a cylinder of raduius 25.0 cm. The moment of inertia about the rotation axis of the reminder of his body is constant and equal to 0.40 kg.m2. If the original angular speed is 0.40 rev/s, What is his final angular speed?

[terryg752]

"hinged as shown"

Where is it shown?

[terryg752]

I will answer the first part, assuming that it is hinged along the side.

Assuming uniform distribution of mass, this is equivalent to the torque of a rod of the same length and same mass. Let the rod lie along the X-axis, hinged at the origin.

Suppose mass be m per unit length.

Then weight of element dx at distance x from origin is mgdx and its torque is mgx dx

Integrating, from 0 to .75 (length)

you get 1/2 mgx^2 from 0 to .75

= 1/2 mg .75^2

mass of length .75 = 13

m = 13/.75

Torque = 1/2 (13/.75) g .75^2

= 1/2 (13) (.75) g

[terryg752]

3.

Torque = 18 times 2.40 = 43.2

I = 2100

Torque = I times angular acceleration

43.2 = 2100 times angular acceleration

Angular acceleration = a = 43.2/2100 = .02057

Angular angular speed after 15 sec = v = 15 times 43.2/2100 = .308571

Angle moved = v^2/2a = .308571^2/(2 times .02057) = 2.314285714

Work done = Torque times Angle moved

= 43.2 times 2.314285714

= 99.97714286

Average Angular Velocity = Angle moved / time

= 2.314285714/15

= 0.154285714

Average Power = Average Angular Velocity times Torque

= 0.154285714 times 43.2

= 6.665142857

[terryg752]

For Part 2

Please provide diagram or full description of the direction of forces

Are the forces in the same plane?

Show direction of 40 degree angle.

[terryg752]

4.

You mean when outstretched : diameter = 1.8 m, Radius = .9

OUTSTRETCHED

Moment of inertia of hands and arms (thin-walled, hollow cylinder)

= Mr^2 = 8 times (.9)^2

= 6.48

Total Moment of Inertia = 6.48 + .40 = 6.88

WRAPPED

Radius = .25 m

Moment of inertia of hands and arms ( cylinder)

= 1/2 MR^2

= 1/2 times 8 times .25^2

= .25

Total Moment of Inertia =.25+ .40 = .65


Initial Angular velocity =.. 40 revolution per sec
= 2 pi times .40 radians per sec

= 2.513274123

L = Angular Momentum = 6.88 times 2.513274123 = 17.29132597


Final Angular Velocity = v

L = 17.29132597 = I times v = .65 v

v = 17.29132597 /.65 = 26.60203995 radians per second

=4.233846154 revolutions per sec