Angular Momentum of a counterweight

[Sheneron]

[SOLVED] Angular Momentum

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.

Homework Equations

$$L = r X P$$
$$\tau = r X F$$
$$L = I\omega$$

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

 

[Doc Al]

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?

 

[Sheneron]

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]

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?

 

[Sheneron]

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

 

[Sheneron]

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

 

[Sheneron]

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.

Homework Equations

$$L = r X P$$
$$\tau = r X F$$
$$L = I\omega$$

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

 

[Doc Al]

Looks good to me.

 

[Sheneron]

Can you help me with part c?

 

[Doc Al]

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).

 

[Sheneron]

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]

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.

 

[Sheneron]

haha yeah i just realized that.

Thank ye

 

[FailWhale42]

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 ?

 

[FailWhale42]

or would it be in N*s/kg?

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

 

[Doc Al]

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.

 

[FailWhale42]

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* :)