Question

Question 1:

A uniform disk with mass 35.4 kg and radius 0.240 m is pivoted at its center about a horizontal, frictionless axle that is stationary. The disk is initially at rest, and then a constant force 27.5 N is applied tangent to the rim of the disk.

Part A

What is the magnitude v of the tangential velocity of a point on the rim of the disk after the disk has turned through 0.260 revolution?

Part B

What is the magnitude a of the resultant acceleration of a point on the rim of the disk after the disk has turned through 0.260 revolution?

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Question 2:

A woman with a mass of 55.0 kg is standing on the rim of a large disk that is rotating at an angular velocity of 0.540 rev/s about an axis through its center. The disk has a mass of 108 kg and a radius of 4.50 m .

Part A

Calculate the magnitude of the total angular momentum of the woman-plus-disk system. (Assume that you can treat the woman as a point.)

Answer 1

here,

1)

mass of disk , m = 35.4 kg

radius of disk , r = 0.24 m

the constant force , f = 27.5 N

A)

let the angular accelration be a

f*r = 0.5 * m*r^2 * alpha

27.5 = 0.5 * 35.4 * 0.24 * alpha

alpha = 6.47 rad/s^2

theta = 0.26 rev = 1.63 rad

let the final speed be w

w = sqrt(2*alpha*theta)

w = sqrt( 2* 6.47 * 1.63)

w = 4.6 rad/s

v = r*w

v = 1.1 m/s

the magnitude v of the tangential velocity of a point on the rim is 1.1 m/s

B)

a = r*alpha

a = 1.55 m/s^2

accelration due to centripital force , a = v^2/r

a = 1.1^2/0.24

a = 5.04 m/s^2

the resultant accelration , a = sqrt( 1.55^2 + 5.04^2)

a = 5.27 m/s^2

the resultant accelration is 5.27 m/s^2

2)

mass of disk , md = 108 kg

radius of disk , r = 4.5 m

mass of woman , mw = 55 kg

angular velocity , w = 0.54 rev/s

w = 3.39 rad/s

the angular momentum , L = I*w

L = ( 0.5 * md *r^2 + mw*r^2) * w

L = ( 0.5 * 108 * 4.5^2 + 55 * 4.5^2) * 3.39

L= 7482.58 kg.m^2.rad/s

the magnitude of the total angular momentum of the woman-plus-disk system is 7482.58 kg.m^2.rad/s