Question

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A uniform solid disk with radius 9 cm has mass 0.5 kg (moment of inertia I = ½MR2). A constant force 12 N is applied as shown. At the instant shown, the angular velocity of the disk is 40 radians/s in the −z direction (where +x is to the right, +y is up, and +z is out of the page, toward you).

The length of the string d is 13 cm. At this instant, what are the magnitude and direction of the angular momentum about the center of the disk? LCM = kg · m2/s Direction: Correct: Your answer is correct.

What are the magnitude and direction of the torque on the disk, about the center of mass of the disk? |τCM| = N · m Direction: Correct: Your answer is correct. The string is pulled for 0.3 s.

What are the magnitude and direction of the angular impulse τCMΔt applied to the disk during this time? |τCM|Δt = N · m · s Direction: Correct: Your answer is correct.

After the torque has been applied for 0.3 s, what are the magnitude and direction of the angular momentum about the center of the disk? kg · m2/s Direction: Correct: Your answer is correct.

At this later time, what are the magnitude and direction of the angular velocity of the disk? rad/s Direction: Correct: Your answer is correct.

Answer 1

1) L = 1/2 *0.5*0.09^2 * 45

L = 0.081 kgm^2/s anticlockwise.

2) Torque = 12 * 0.09 = 1.08 Nm clockwise

3) I = T . t = 1.08* 0.3 = 0.324 Nms ( clockwise )

4) I = change in angular momentum = Ang mom final – original ang mom

Ang mom final = 0.324 – 0.081 = 0.243 kgm^2/s clockwise

5 ) Ang vel = Ang mom/ Moment of inertia =( 0.243) /( 1/2 *0.5* 0.09^2) = 120 rad/s

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