Every rotational motion have torque?

[MechaMZ]

Every rotational motion have torque? Pls come and take a look..

Are torque exsiting at every rotational motion?
how about free fall or wheel rotating without any force act on it??

Question:
A 0.8m diameter, 50kg flywheel with a radius of gyration of 0.283m must be braked from 60 rpm to rest in 2s. calculate the tangential force required to accomplish this braking action.

angular acceralation = -3.142rad/s sq
I = 4.004kgm sq

torque equation on ratational motion of the flywheel is :

Torque - ( T x 0.4m ) = I x (angular acceleration) < T is tangential force >
0 - ( T x 0.4 ) = 4.004 ( -3.142 )
-0.4T = -12.58
T = 31.4 N

why the torque here is equal to zero??

anyone can help and explain...
ur reply is greatly appreciated... thankzzz

 

[russ_watters]

Clearly, if there are no forces, there are no torques. What this solution is saying to you is that the sum of the torques is equal to zero.

 

[Doc Al]

torque equation on ratational motion of the flywheel is :

Torque - ( T x 0.4m ) = I x (angular acceleration) < T is tangential force >
0 - ( T x 0.4 ) = 4.004 ( -3.142 )
-0.4T = -12.58
T = 31.4 N

why the torque here is equal to zero??

The torque is not zero. That equation should be: (net)Torque = I x (angular acceleration). Presumably, the only force producing a torque on the flywheel is the applied tangential force. Thus: net torque = T x 0.4m (using your notation).

 

[MechaMZ]

Clearly, if there are no forces, there are no torques. What this solution is saying to you is that the sum of the torques is equal to zero.

sorry, one more question..

when/how i know the sum of the torque is equal to zero in the rotational motion?

 

[Doc Al]

when/how i know the sum of the torque is equal to zero in the rotational motion?

Just to be clear: In this problem the sum of the torques is not zero. (If it were, the angular acceleration would be zero.)

 

[MechaMZ]

sum of torque = I x (angular acceleration)

torque1 - torque2 = I x (angular acceleration)

resolved the torque into 2 direction, anti clock'wise or clock'wise.
direction of acceleration is assumed as positive.

torque1 is zero, because no force in this direction.
torque2 is negative sign, because it is opposite direction with acceleration

0 - (tangential force x 0.4) = (4.004)(-3.142)
- (tangential force x 0.4) = -12.580
tangential force = 31.45 N


the sum of torque is negative sign, -12.580Nm. This means there is a resultant torque with anti clock'wise direction. And it is torque2.


pls leave ur comments abt my solution, i will kindly accept, thankz

 

[MechaMZ]

however, i found tht torque1 actually is not exsiting. torque2 is negative just because it's direction is opposite with the angular acceleration force : )

 

[russ_watters]

Just to be clear: In this problem the sum of the torques is not zero. (If it were, the angular acceleration would be zero.)

Sorry about that - the question in the first two sentences is different from what the problem was asking and I didn't look at the problem...