Question

The flywheel in the shape of a solid cylinder of mass 75.0 kg and diameter 1.40 m rotates about an axis passing through the center of mass. a. what constant torque is required to bring it up to an angular speed of 320 rev/min in 8.00 s, starting from rest?

Answer 1

The torque required for the cylinder is 76.99 N m.

What is torque?

Torque may be defined as the product of the force and the distance perpendicular to the force from the axis of rotation.

Calculation of the angular acceleration:

The first kinematic equation for the rotational motion is,

ω=ω₀+αt

where ω₀ is the initial angular velocity, ω is the final angular velocity, α is the angular acceleration and t is the time. Since the cylinder starts from the rest, its initial angular velocity is zero.

Here, ω=320 rev/min or ω=320*2π/60 rad/sec, ω₀=0 and t=8.00 sec. Put these values in the above equation and solve it to get the value of angular acceleration.

320*2π/60=0+α*8

α=320*2π/(8*60)

α=4.19 rad/s^2

Calculation of the torque:

The torque T is given by the formula,

T=Iα

where I is the moment of inertia.

For the cylinder, I=1/2*mr^2

where m is the mass of the cylinder and r is the radius of the cylinder.

So,

T=1/2*mr^2*α

Here m=75 kg, radius r is half of the diameter r=1.4/2=0.7 m, and α=4.19 rad/s^2. Using these values in the formula of torque.

T=1/2*(75)*(0.7)^2*4.19

T= 76.99 N m

Learn more about the torque.

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