# Torque and tension?

A string is wrapped several times around the rim of a small hoop with radius r and mass m. The free end of the string is held in place and the hoop is released from rest.. (picture attached~)

So I tried using moment of inertia as mass*radius^2, so torque would be MR^2*alpha, and it would also be tension*R.

Solving for tension would give me mass*acceleration as tension.

Then net force would be tension-mass*gravity.

m*a=T-m*g
T=T-m*g

Now that doesn't make any sense.

Can anyone help?

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HallsofIvy
Homework Helper
You're right- it doesn't make sense. T is NOT equal to m*a- for one thing, that doesn't take into account the increasing angular momentum.

The moment of inertia you used is for a hoop, with the axis of rotation at the centre of the hoop. You need to use the moment of inertia with the axis of rotation at the outter radius.
In this case:

$$I = \frac{mR^2}{2}$$

Regards,

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Doc Al
Mentor
So I tried using moment of inertia as mass*radius^2, so torque would be MR^2*alpha, and it would also be tension*R.
Nothing wrong with that. You have applied the acceleration constraint that connects $\alpha$ with a: $a = \alpha R$.

Solving for tension would give me mass*acceleration as tension.
Right. T = ma (where "a" is the magnitude of the acceleration).

Then net force would be tension-mass*gravity.
Right. The net force on the hoop is T - mg, taking up as positive. Applying Newton's 2nd law gives you: T - mg = - ma. (note the minus sign; the acceleration acts downward thus is negative)

m*a=T-m*g
T=T-m*g

Now that doesn't make any sense.
It doesn't make sense because you mixed up the signs. Combine these equations and it will make sense:
T - mg = - ma
T = ma

heres what happened when I did it:
$$RT = I\alpha$$
$$RT = I\frac{a}{R}$$

since the moment of inertia is $$\frac{mR^2}{2}$$

then:

$$RT = \frac{mR^2}{2} \frac{a}{R}$$
$$T = \frac{ma}{2}$$

now we can sub this in into our free body diagram equation:

You should be able to do the rest.

Regards,

Doc Al
Mentor
The moment of inertia you used is for a hoop, with the axis of rotation at the centre of the hoop. You need to use the moment of inertia with the axis of rotation at the outter radius.
In this case:

$$I = \frac{mR^2}{2}$$
There's nothing wrong with taking a point on the edge of the hoop as your axis, but that's not the correct moment of inertia.

$$T - mg = -ma$$ because acceleration is downward.

$$2T = mg$$

$$T = (mg)/2$$

Darn little technicalities.

Thanks guys! You helped me finish my homework on time .

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