Rotational Dynamics of a Motor-Driven Drum System

[Dan350]

I think I on the rigth track, would you revise it please

1. A motor is connected to a solid cylindrical drum with a radius of 0.6m and a mass of 52kg. A massless rope is attached to the drum and tied at the other end to a 38kg weight so the rope will wind onto the drum as it turns. The weigth is rising with an acceleration of 1.5m^2. See Image below.

a) What is the angular acceleration of the drum?

b) At the moment that the drum is rotating at ω=3.5rad/s what is the rate at which the motor is doing work?


2. Homework Equations

T=rfsinθ
T=Iα
F=ma
I=1/2mr^2
)



3. The Attempt at a Solution

For part a:
F=ma
Fapp-mg=ma
Fapp=m(a+g)
So I got the force of the acting on the system to use it in my torque so
T=rfsin(90)
T=257.64N
then we get the intertia of the drum, so I=1/2(52kg)(.6m^2)= 9.36kgm^2

so we can apply it
T/I=α

∴ 257.64/9.36= 27.52rad/s^2

attempt at b

W=∫Tdθ from θi to θf
∴
1/2Iωf^2-1/2Iωi^2

so plugin in values
1/2(9.36kgm^2)(3.5rad/s)^2
=57.33N

Am I correct? if not, would you tell me why and where am I wrong please.

Thank You
-Dan

 

[ehild]

The rope is wound on the rim of the drum. The linear speed of the perimeter of the drum is the same as the speed of rise of the mass, as the string does not stretch. If the speeds are the same, so are the accelerations. And how are the linear speed/ acceleration related to angular speed/acceleration? So what is the angular acceleration of the drum?

You determined the force of tension as m(a+g) that is needed to accelerate the hanging mass upward. If the motor exerts the torque equal to m(a+g)R the system will not accelerate. Accelerating the whole system, you need extra torque from the motor. How much is the torque of the motor then?

Note that you have to determine the power of the motor (instead of force). How do you calculate rotational work/power?

ehild