Rotation of a drum attached to a spring

[Datta108]

Homework Statement

Rotation of a small drum of mass moment of inertia 0.44kg.m^2 and with frictional resistance torque 0.3 N.m is initiated by a string wound around its 20 mm diameter shaft and attached to a stretched spring, as shown in the figure. If the spring, which has a stiffness 2N/mm is initially stretched by 120 mm before the drum is released from rest, determine the maximum angular velocity, in rpm, attained by the drum.

Homework Equations

Tnet = Iα
SE = 1/2 kx^2
KE of drum = 1/2Iw^2

The Attempt at a Solution

The spring energy provided by the stretched spring is given to the drum and converted to the kinetic energy of the drum. Equating these two gives w = 23.4 rad/s^2. I don't know how to proceed from here. I believe we have to find the T provided by the spring. Then, the first equation can be used to find the difference in T and therefore alpha can be found. From this v can also be found. However, I am stuck at finding T provided by spring.

Any help (ASAP) will be greatly appreciated.

Many Thanks!

 

[haruspex]

Equating these two gives w = 23.4 rad/s^2

That is not going to help since the peak rate isprobably reached before the spring becomes slack.

stuck at finding T provided by spring

There is no difficulty finding the initial tension, is there?
Consider when the spring has contracted by a distance x from its initial stretch. What work has been done by the spring? What is the KE of the drum?

 

[scottdave]

Using the frictional torque and the radius, you can find an effective "frictional force" acting at the edge.

 

[Datta108]

That is not going to help since the peak rate isprobably reached before the spring becomes slack.

There is no difficulty finding the initial tension, is there?
Consider when the spring has contracted by a distance x from its initial stretch. What work has been done by the spring? What is the KE of the drum?

Sorry, did you mean torque?

Work done by the spring would be 1/2kx^2.
Won't the KE of the drum equal the SE of the spring?Sorry if I am asking stupid questions here, but my tutor did not go over such problems in detail.
(This is NOT to complete my homework by the way. I am studying for an exam that's coming up)

 

[Datta108]

Using the frictional torque and the radius, you can find an effective "frictional force" acting at the edge.

Do you mean 0.3 N.m * 0.010 m?

 

[scottdave]

Look at the units. You want to divide (0.3 N*m) / (0.01 m) to get 30 Newtons of friction, which works against motion of the drum (so it is in opposite direction of spring force).

 

[haruspex]

Work done by the spring would be 1/2kx^2.

No. It started with extension L (=0.12m) and I defined x as the reduction in extension, so:
- what PE did the spring start with? (Use L, not the numerical value.)
- what PE does it still have?
- what work has it done?

Won't the KE of the drum equal the SE of the spring?

No, there s friction. How much work has been done against that?

 

[Datta108]

No. It started with extension L (=0.12m) and I defined x as the reduction in extension, so:
- what PE did the spring start with? (Use L, not the numerical value.)
- what PE does it still have?
- what work has it done?

No, there s friction. How much work has been done against that?

Sorry, I am not able to think about this problem. Could you please give me a step by step method for this question?
I have a mechanics exam tomorrow. Thank you.

 

[gneill]

Sorry, I am not able to think about this problem. Could you please give me a step by step method for this question?
I have a mechanics exam tomorrow. Thank you.

Sorry, helpers are not permitted to provide step by step solutions for homework problems. They can offer hints and point out errors in work you've shown, or offer suggestions concepts to think about or try.

 

[Datta108]

Sorry, helpers are not permitted to provide step by step solutions for homework problems. They can offer hints and point out errors in work you've shown, or offer suggestions concepts to think about or try.

Sorry, I was not aware of this. Thank you all!