# Angular acceleration of a massless pulley

## Main Question or Discussion Point

Consider a massless string going over a massless frictionless pulley with masses M1 and M2 at the end of the string.

The tension in the string would be same.

The torque equation of the pulley says (T1-T2)=Iα .Now L.H.S is zero since T1=T2 and also I=0 .

So,we have a condition 0=(0)(α) which makes α indeterminate .But we know that the pulley rotates with some angular acceleration.

So ,how is α determined ?

Is α=a/R where R is the radius of the pulley and a=[(M2-M1)g]/(M1+M2) ?

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If there's no friction, a massless pulley will just stay in place as the rope slides along it.

By frictionless,I mean frictionless at the axle .There is sufficient friction present between the pulley and the string.
It is a different matter although that a massless pulley requires zero torque(zero friction) to rotate.

The pulley will rotate .But from the torque equation , we cant determine the angular acceleration.

• 1 person
Thanks dauto :)

Another clarification i am looking at is whether the net torque on the pulley determines its angular acceleration ,or is it the other way round ,i.e angular acceleration determines the torque ?

When the pulley has mass , (T2-T1)R = Iα and α=a/R .

Does that mean the linear acceleration of the masses 'a' determines the angular acceleration 'α' .This 'α' in turn determines the value of net torque ?

Chestermiller
Mentor
Thanks dauto :)

Another clarification i am looking at is whether the net torque on the pulley determines its angular acceleration ,or is it the other way round ,i.e angular acceleration determines the torque ?

When the pulley has mass , (T2-T1)R = Iα and α=a/R .

Does that mean the linear acceleration of the masses 'a' determines the angular acceleration 'α' .This 'α' in turn determines the value of net torque ?
In this case, the force balances on the masses and the torque balance on the pulley need to be coupled with one another to solve for the angular acceleration of the pulley or, equivalently, the linear accelerations of the masses (using α=a/R). Write the three equations out, and see how this plays out.

Chet

• 1 person
Thanks Chet....I understand what you are saying...I am just unsure about the case of a massless pulley...

In the case of the massless pulley ,the angular acceleration is solely determined by the linear acceleration of the masses .Right?

Chestermiller
Mentor
Thanks Chet....I understand what you are saying...I am just unsure about the case of a massless pulley...

In the case of the massless pulley ,the angular acceleration is solely determined by the linear acceleration of the masses .Right?

(T1-T2)R=Iα=Ia/R
M1a=M1g-T1
M2a=T2-M2g
So, combining these equations, we get:
$$a(M_1+M_2+\frac{I}{R^2})=(M_1-M_2)g$$
When I = 0, you get the linear acceleration result.

• 1 person
Thanks Chet 