# Angular Momentum of a counterweight

[SOLVED] Angular Momentum

1. Homework Statement
A 4.60 kg counterweight is attached to a light cord, which is wound around a spool (refer to Fig. 10.20). The spool is a uniform solid cylinder of radius 8.00 cm and mass 1.30 kg.
(a) What is the net torque on the system about the point O?
Magnitude
(b) When the counterweight has a speed v, the pulley has an angular speed = v/R. Determine the total angular momentum of the system about O.
(c) Using the fact that = dL/dt and your result from (b), calculate the acceleration of the counterweight.
2. Homework Equations
$$L = r X P$$
$$\tau = r X F$$
$$L = I\omega$$

3. The Attempt at a Solution
I was able to solve the first part:
$$\tau = rF$$
$$\tau = (0.08)(4.6)(9.8) = 3.61 Nm$$

I am not sure how to solve part B though, here is what I tried but it was wrong:

$$L_{net} = rp + I\omega$$
$$L_{net} = r X mv + I*\frac{v}{r}$$
$$L_{net} = v (r X m + \frac{I}{r})$$

I factored out the v because it doesn't give you the velocity, and the answer space looks like this:

________ X v kg *m^2/s

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Doc Al
Mentor
I am not sure how to solve part B though, here is what I tried but it was wrong:

$$L_{net} = rp + I\omega$$
$$L_{net} = r X mv + I*\frac{v}{r}$$
$$L_{net} = v (r X m + \frac{I}{r})$$
Looks OK to me--keep going.

Hint: What's the rotational inertia of a solid cylinder?

Rotational inertia of a solid cylinder is 0.5MR^2

With my equation there, at the rXm, do I have to do the cross product?

Doc Al
Mentor
Rotational inertia of a solid cylinder is 0.5MR^2
Good.

With my equation there, at the rXm, do I have to do the cross product?
You can only take cross products of vectors, so r X m doesn't make sense. (I was wondering if you meant that X as a cross product or just as multiplication.) The time to do the cross product is before you factor out the speed.

In the definition of angular momentum of a particle, $\vec{L} = \vec{r}\times m\vec{v}$, $\vec{r}$ is the position vector of the particle. The cross product turns out to be easy to evaluate in terms of r (the radius). So what is the angular momentum of the particle?

I don't know. I am a little unclear about how to do cross products.

Can someone tell me how I would go about cross producting $\vec{L} = \vec{r}\times m\vec{v}$

1. Homework Statement
A 4.60 kg counterweight is attached to a light cord, which is wound around a spool (refer to Fig. 10.20). The spool is a uniform solid cylinder of radius 8.00 cm and mass 1.30 kg.
(a) What is the net torque on the system about the point O?
Magnitude
(b) When the counterweight has a speed v, the pulley has an angular speed = v/R. Determine the total angular momentum of the system about O.
(c) Using the fact that = dL/dt and your result from (b), calculate the acceleration of the counterweight.

2. Homework Equations
$$L = r X P$$
$$\tau = r X F$$
$$L = I\omega$$
3. The Attempt at a Solution
I was able to solve the first part:
$$\tau = rF$$
$$\tau = (0.08)(4.6)(9.8) = 3.61 Nm$$

I am not sure how to solve part B though,

$$L_{net} = rp + I\omega$$
$$L_{net} = (r)mv + I*\frac{v}{r}$$

$$L_{net} = v (rm + \frac{I}{r})$$
$$L_{net} = v [ (.08)(4.6) + \frac{1}{2}(1.30)(.08)]}$$
$$L_{net} = v(0.42)$$

I factored out the v because it doesn't give you the velocity, and the answer space looks like this:

________ X v kg *m^2/s

Is that how I would go about part B? Thanks

Last edited by a moderator:
Doc Al
Mentor
Looks good to me.

Can you help me with part c?

Doc Al
Mentor
Can you help me with part c?
What's the full statement of part c? (Your initial post has something missing.) You'll need to take the derivative of L (that's what dL/dt means).

sorry, yes I know i have to take the derivative...
Here is the post:
Using the fact that t = dL/dt and your result from (b), calculate the acceleration of the counterweight.

so would it be: 3.61 = dL/dt and dL/dt= 0.42dv/dt and dv/dt is acceleration so would it be a= 3.61*0.42

Doc Al
Mentor
so would it be: 3.61 = dL/dt and dL/dt= 0.42dv/dt and dv/dt is acceleration
Good.
so would it be a= 3.61*0.42
Divide, not multiply, by 0.42.

haha yeah i just realized that.

Thank ye

Re: [SOLVED] Angular Momentum

just wondering what units the final answer is in - if the counterweight is just accelerating downwards, do we still write the answer in rads/sec^2 ?

Re: [SOLVED] Angular Momentum

or would it be in N*s/kg?

(N*m)/(kg*m^2/s)

Doc Al
Mentor
Re: [SOLVED] Angular Momentum

just wondering what units the final answer is in - if the counterweight is just accelerating downwards, do we still write the answer in rads/sec^2 ?
The final answer for what question? rads/sec^2 are units for angular acceleration; the counterweight has linear acceleration, which has units of m/s^2.

Re: [SOLVED] Angular Momentum

The final answer for what question? rads/sec^2 are units for angular acceleration; the counterweight has linear acceleration, which has units of m/s^2.
thankyou! :) i was thinking angular acceleration *facepalm* :)