An electric motor (r_m = 0.05m) turns a flywheel through a drive belt that joins

Question

An electric motor (r_m = 0.05m) turns a flywheel through a drive belt that joins a pulley on the motor to a pulley that is rigidly attached to the flywheel as shown in the figure below. The flywheel is a solid disk with a mass of 58.5 kg and a radius R = 0.625 m. It turns on a frictionless axle. Its pulley has a mass that can be ignored and a radius of r = 0.130 m. a.) The relationship between the angular acceleration of the motor and the wheel is r_m/r alpha_omega = alpha_m. Use this to determine the angular acceleration of the wheel if the motor has an angular acceleration of 1.67rad/s^2 b.) After 10s, what is the angular velocity of the wheel? c.) At this instant the belt snaps and we wish to bring this wheel to a stop. What net torque is required to stop this wheel within 5s? d.) Through how many revolutions does this wheel turn before coming to a stop?

Explanation / Answer

rm = radius of the motor = 0.05m

r = radius of the pulley = 0.13m

R = radisu of the flywheel = 0.625m

M = mass of the flywheel = 58.5kg

m = angular accelration of the motor

w = angular accelration of the flywheel

a) (rm/r)w = m

or w =rm/rm

we have m = 1.67rad/s2, So,

w = (0.13m)(1.67rad/s2)/0.05m = 4.342rad/s2

So angular acceleration of the wheel is w = 4.342rad/s2

b) we use the formula

f = i + wt ------------- (1)

where f = final angular velocity of the wheel, and i = initial angular velocity of the wheel

initial angular velocity of the wheel is zero.So , the equation (1) becomes

f = 0 + (4.342rad/s2)(10s)

or f = 43.42rad/s

So angula velocity of the wheel after 10s is f = 43.42rad/s

(c) The belt snaps after 10s. At this instant the angula velocity of the wheel is = 43.42rad/s.

Let us assume the angular acceleration provided by the torque be . In this case the final angular velocity would be zero and the initial angular velocity is = 43.42rad/s. We use the following formula:

f = i + t

or 0 = 43.42rad/s + (5s)

or = -8.684rad/s2, is -ve as it is deceleration.

So torque needed is T = I

where I = moment of inertia of the wheel = MR2/2 = (58.5kg)(0.625m)2/2 = 11.4257kg-m2

So T = (11.4257kg-m2)(-8.684rad/s2) = -99.22N-m is the net torque needed.

-ve sign of the torque indicates that it has to be applied in a direction opposite to the initial acceleration.

(d) We use the formula:

f2 = i2 + 2 ---------- (2)

where f = final angular velocity of the wheel = 0, in this case

and i = initial angular velocity of the wheel = 43.42rad/s

= angular displacement from the time when torque is applied to stop the wheel to the time when it stops.

So using (2) we get:

0 = (43.42rad/s)2 - 2(8.684rad/s2)

or = 108.55radian

Now one revolution = 2 radian = 6.28radian

So required number of revolutions = (108.55radian)/6.28radian = 17.285

So number of total complete revolutions will be 17.

This concludes the answers. Check the answer and let me know if it's correct. If you need anymore clarification I will be happy to oblige....

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