# Thread: Linear and angular acceleration of flywheel

1. ## 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!?

2. ## Re: Linear and angular acceleration of flywheel

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!?
$m$ = hanging mass

$d$ = linear displacement of the hanging mass

$M$ = wheel mass

$R$ = wheel radius

$r_g$ = radius of gyration

(a) linear kinematics ...

$d = \frac{1}{2}at^2$

$a = \frac{2d}{t^2}$

(b) assuming no slippage ...

$\alpha = \frac{a}{R}$

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

$mg - T = ma$

$T = m(g-a)$

(d) $\tau_{net} = I \alpha$

$TR - \tau_f = M(r_g)^2 \alpha$

solve for the frictional torque, $\tau_f$

3. ## Re: Linear and angular acceleration of flywheel

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?

4. ## Re: Linear and angular acceleration of flywheel

(d) torque caused by the hanging mass , $\tau = T \cdot R$ , where $T$ is the tension in the cord attached to the hanging mass

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

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

$T \cdot R - \tau_f = I \alpha$

solve for $\tau_f$ ...

$\tau_f = T \cdot R - I \alpha$

finish it.

5. ## Re: Linear and angular acceleration of flywheel

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

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