A solid disk of mass equal to 7.0 kg and radius of 30 cm changes its rotations g

Question

A solid disk of mass equal to 7.0 kg and radius of 30 cm changes its rotations gradually from 45 rpm to 120 rpm in 5.0 sec. find the angular acceleration in rad/s^2. how many revolutions has the disk turned in this 5.0 seconds? find the change in rotational kinetic energy. find its moment of inertia in kgm^2. find the magnitude of torque needed to make such a change in rpm. find the tangential acceleration of a point P on the outer rim of the disk. The fan blade in a jet engine rotates from rest with the angular acceleration alpha = 20t^2. find the angular velocity as a function of time. find the angular displacement as a function of time. if the rotational inertia of the blade is 5.0 kgm^2, find the work done on the blade in the first initial 3.0 seconds. A hollow sphere (M = 3.0 kg and R = 60 cm) is rolling down an inclined surface with a height of 2.0 meters and an inclined angle of 25 degrees. find the moment of inertia of the sphere, in kgm^2. find the speed of the center of mass when it reaches the bottom of the hill.

Explanation / Answer

mass, m=7kg

radius, r=30cm

angular speed, w1=45 rev/min , w2=120 rev/min

time, t=5sec

3)

acceleration, alpa=(w2-w1)/t

alpa=(120-45)*(2pi/60)/(5)

alpa=1.57 rad/sec^2

4)

w2^2-w1^2=2*alpa*theta

(120*(2pi/60))^2 - (45*(2pi/60))^2 = 2*1.57*theta

==> theta=43.22 degrees

n=(43.22/360) rev

n=0.12 rev

6)

moment of inertia of solid disk, I=1/2*m*r^2

I=1/2*7*(30*10^-2)^2

I=0.315 kg.m^2

5)

change in K.E =K2-K1

=1/2*I*w2^2 - 1/2*I*w1^2

=1/2*0.315*((120*(2pi/60))^2 - (45*(2pi/60))^2)

=21.37 J

7)

torque=I*alpa

=0.315*1.57

=0.494 N.m

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