A uniform disc, which has a mass of 3.50 kg and a radius of 28.0 cm, is accelera

Question

A uniform disc, which has a mass of 3.50 kg and a radius of 28.0 cm, is accelerated from rest by a tangential force of 28.0 N applied to the outer edge of the disc. 1. a. b. c. d. e. f. g. h. What is the moment of inertia of this disc? What is the magnitude of the torque being applied to this disc? What will be the angular acceleration of this disc? What will be the angular velocity of this disc after 8.0 seconds? What will be the linear velocity of a point on the outer edge of the disc at the end of 8.0s? What will be the angular displacement of this disc during the 8.0 s period? What will be the linear displacement of a point on the outer edge of this disc during the 8 s? How much work was done on this disc during the 8.0 s time period? what will be the final angular momentum of this disc? What is the final angular kinetic energy of this disc? j.

Explanation / Answer

1.

Moment of inertia of disc is

I=MR2/2

I=3.5*0.282/2

=0.1372

b)

Magnitude of torque is

Torque= Force*R

=28*0.28

=7.84

c)angular accelration is

torque= accleration*I

So,

=7.84/0.1372

=57.14285

d.

Angular velocity after 8 seconds will be,

=57.14285*8

=457.14285

e.

linear velocity will be,

=457.14285*0.28(angular velocity*radius)

=128m

f.

angular displacment will be,

=0.5*57.14285*82

=1828.5712 radians.

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