Thread: Linear and angular acceleration of flywheel

Hi, I have the following question and was wondering if anyone could help:

A mass of 0.5kg is suspended from a flywheel, of mass 3kg, outside radius 0.3m and radius of gyration 0.212m. If the mass is release from rest and falls a distance of 0.5m in 1.5 seconds, calculate:

(a) The linear acceleration of the mass
(b) The angular acceleration of the wheel
(c) The tension in the rope
(d) The frictional torque resisting the motion

Equations (I think!); a = g[(2m/(M+2m)] v=s/d w=v/r T=mg[M/(M+2m)] I=mk2

I think (although I could be wrong) that (a) can be calculated using a = g[(2m/(M+2m)] where m = 0.5kg, M = 3kg and g = 9.81m/sec2. This would give an answer of 2.45m/sec2, unless I'm being stupid and its as straight forward as acceleration due to gravity = 9.81m/sec2?

For (b) Angular velocity = linear velocity/radius = (0.5/1.5)/0.3 = 1 rad/sec

Therefore, angular acceleration = change in angular velocity/time = (1-0)/1.5 = 0.6 rad/sec2

(c) Does T, tension=mg[M/(M+2m)] = 3.78 N ?

(d) I'm not sure how to approach this one, I know frictional torque = total torque - accelerating torque and moment of inertia and radius of gyration must be included at some point but haven't solved it yet - help!?

Originally Posted by chunkylumber111

Hi, I have the following question and was wondering if anyone could help:

A mass of 0.5kg is suspended from a flywheel, of mass 3kg, outside radius 0.3m and radius of gyration 0.212m. If the mass is release from rest and falls a distance of 0.5m in 1.5 seconds, calculate:

(a) The linear acceleration of the mass
(b) The angular acceleration of the wheel
(c) The tension in the rope
(d) The frictional torque resisting the motion

Equations (I think!); a = g[(2m/(M+2m)] v=s/d w=v/r T=mg[M/(M+2m)] I=mk2

I think (although I could be wrong) that (a) can be calculated using a = g[(2m/(M+2m)] where m = 0.5kg, M = 3kg and g = 9.81m/sec2. This would give an answer of 2.45m/sec2, unless I'm being stupid and its as straight forward as acceleration due to gravity = 9.81m/sec2?

For (b) Angular velocity = linear velocity/radius = (0.5/1.5)/0.3 = 1 rad/sec

Therefore, angular acceleration = change in angular velocity/time = (1-0)/1.5 = 0.6 rad/sec2

(c) Does T, tension=mg[M/(M+2m)] = 3.78 N ?

(d) I'm not sure how to approach this one, I know frictional torque = total torque - accelerating torque and moment of inertia and radius of gyration must be included at some point but haven't solved it yet - help!?

= hanging mass

= linear displacement of the hanging mass

= wheel mass

= wheel radius

= radius of gyration

(a) linear kinematics ...





(b) assuming no slippage ...



(c) Newton's 2nd law ...





(d)



solve for the frictional torque,

That's great, thanks a lot for your help. I think I've got some answers now but I'm still not 100% sure on how (d) is worked out, could you give me a little more help please?

So (a) = 0.44m/s^2

(b) = 0.44/0.3 = 1.47 m/s^2

(c) 0.5(9.81-0.44) = 4.69N

(d) Is the T in (TR - Tf) the tension in the rope worked out from part (c)?

or is it Tnet = Ia and since =M(k^2) where a = alpha and k = radius of gyration

Therefore Tnet = M(k^2)a = 3x(0.212^2)x(1.47) = 0.198 Nm

in which case Tf = (0.198x0.3) - 3(0.212^2)x1.47 = -0.139Nm ?

Is the above correct?

(d) torque caused by the hanging mass , , where is the tension in the cord attached to the hanging mass

torque due to friction in the axle (what you want to find) ...

net torque is the difference between the torque generated by the hanging mass and the frictional torque ...



solve for ...



finish it.

Ok now I get it! So Tf = 1.407 - 0.198 = 1.209Nm? Thanks