If tension provides the torque to this pulley, what is its equation?

[haseebaslam]

Homework Statement

A bucket of weight 15.0 N (mass of 1.53 kg) is hanging from a cord wrapped around a pulley. The pulley has a moment of inertia of pulley=0.385 kgm^2 (of radius R = 33.0 cm). The cord is not stretched nor slip on the pulley. The pulley is observed to accelerate uniformly. If there is a frictional torque at the axle equal to, =1.10Nm. First calculate the angular acceleration, α, of the pulley and the linear acceleration of the bucket.

Relevant Equations

a = α*radius
Net Torque = I*angular acceleration

15 - Tension = 1.53*a
T = 15 - 1.53*R*α
T*0.33 - 1.10 = Iα
Is this approach correct?

 

[topsquark]

Homework Statement: A bucket of weight 15.0 N (mass of 1.53 kg) is hanging from a cord wrapped around a pulley. The pulley has a moment of inertia of pulley=0.385,m^2 (of radius R = 33.0 cm). The cord is not stretched nor slip on the pulley. The pulley is observed to accelerate uniformly. If there is a frictional torque at the axle equal to, =1.10⋅m. First calculate the angular acceleration, α, of the pulley and the linear acceleration of the bucket.
Relevant Equations: a = α*radius
Net Torque = I*angular acceleration

15 - Tension = 1.53*α
T = 15 - 1.53*r*α
T*0.33 - 1.10 = Iα
Is this approach correct?

Always start with an extended FBD...

You have
$15 - T = 1.53 a$

and
$T = 15 - 1.53ra$

Is the "a" in this second equation supposed to be an $\alpha$?

-Dan

 

[haruspex]

Some problems with the units. Is the MoI of the pulley 0.385 kg m²?
Is the frictional torque 1.10Nm?
Your first equation has force on the left, mass/time² on the right, but I see you corrected that in the next line.

But your approach looks right.

 

[haseebaslam]

Always start with an extended FBD...

You have
$15 - T = 1.53 a$

and
$T = 15 - 1.53ra$

Is the "a" in this second equation supposed to be an $\alpha$?

-Dan

Sorry about that, the a in my first equation is for linear acceleration. In the second equation, I substitute linear acceleration to get angular acceleration in the equation.

 

[kuruman]

Sorry about that, the a in my first equation is for linear acceleration. In the second equation, I substitute linear acceleration to get angular acceleration in the equation.

Angular acceleration is $\alpha$ (alpha) and the relation is $a=\alpha r.$ Do you see the difference between the symbols for Greek $\alpha$ and Latin $a$?

So if you wish to substitute, you should write $T=15-1.53r\alpha$, not $T=15-1.53ra.$

 

[Orodruin]

Note: I highly suggest working with symbols rather than values until you reach a final expression for whatever you wish to compute. Inserting numbers at an early stage makes it less clear where things come from and makes it harder to check for errors. You also run the risk of introducing rounding errors down the line. Only insert numerical values at the very end.