What is the final speed of a block and the angular speed of a pulley?

[pcandrepair]

[SOLVED] Speed of a Block

Homework Statement

The sliding block has a mass of 0.830 kg, the counterweight has a mass of 0.430 kg, and the pulley is a hollow cylinder with a mass of 0.350 kg, an inner radius of 0.020 m, and an outer radius of 0.030 m. The coefficient of kinetic friction between the block and the horizontal surface is 0.250. The pulley turns without friction on its axle. The light cord does not stretch and does not slip on the pulley. The block has a velocity of 0.820 m/s toward the pulley when it passes through a photogate.

(a) Use energy methods to predict its speed after it has moved to a second photogate, 0.700 m away.
_________ m/s

(b) Find the angular speed of the pulley at the same moment.
_________ rad/s

Homework Equations

Moment of Inertia for a hollow sphere: Icm = 1/2 M (R1^2 + R2^2)

Friction = \mu*M1*g*diameter

The Attempt at a Solution

For part A i used the following equation to solve for the final velocity:

1/2(M1*Vi^2) + 1/2(M2*Vi^2) - friction = 1/2(M1*Vf^2) + 1/2(M2*Vf^2) + M2*g*h


1/2(.83Kg * (.82m/s^2)) + 1/2(.43Kg * (.82m/s^2)) - (.25)*(.83Kg)*(9.8)*(.7m) = 1/2(.83Kg * Vf^2) + 1/2(.43Kg * Vf^2) + (.83Kg * 9.8 * .7m)

-1 = 1/2 Vf^2 (.83Kg + .43Kg) + 2.9498
Vf^2 = -4.5371
Vf = 2.13 m/s

This answer is incorrect. I cannot figure out where I went wrong. If someone could help me out, it would be much appreciated. Thanks!

I also tried part b with this number. I'm pretty sure this is the correct formula but I need the right number from part a to get the correct answer.

V = R\omega
\omega = V/R
\omega = (2.13m/s) / (sqrt((.02^2) + (.03^2)))
\omega = 59.0756 radians/sec

 

[Doc Al]

For part A i used the following equation to solve for the final velocity:

1/2(M1*Vi^2) + 1/2(M2*Vi^2) - friction = 1/2(M1*Vf^2) + 1/2(M2*Vf^2) + M2*g*h


1/2(.83Kg * (.82m/s^2)) + 1/2(.43Kg * (.82m/s^2)) - (.25)*(.83Kg)*(9.8)*(.7m) = 1/2(.83Kg * Vf^2) + 1/2(.43Kg * Vf^2) + (.83Kg * 9.8 * .7m)

In calculating the change in PE, note that the hanging mass lowers. The change in height should be negative.

 

[pcandrepair]

When I use a negative height, I get 1.75888 m/s and it tells me my answer is wrong but within 10% of the correct value.

 

[Doc Al]

You also forgot the rotational KE of the pulley.

 

[pcandrepair]

I added 1/2 Iw^2 to each side using the initail veloctiy for w on the left side and w on the right side is the same as the Vf on the right side? When I do this I'm getting 2.397m/s which wouldn't be in the 10%.

 

[Doc Al]

I added 1/2 Iw^2 to each side using the initail veloctiy for w on the left side and w on the right side is the same as the Vf on the right side?

\omega = v/r, where r is the outer radius.

 

[pcandrepair]

Now I'm getting 1.75919 m/s. This seems way to close to my answer from post #3.

 

[Doc Al]

If you spell out the entire equation you are using (like you did in post #3) and show the intermediate values you get, I'll look it over.

 

[pcandrepair]

1/2(M1*Vi^2) + 1/2(M2*Vi^2) - friction + 1/2 (I)(V/R)^2 = 1/2(M1*Vf^2) + 1/2(M2*Vf^2) + M2*g*h + 1/2 (I)(V/R)^2

I found (I) to equal .000228

1/2(.83Kg * (.82m/s^2)) + 1/2(.43Kg * (.82m/s^2)) - (.25)*(.83Kg)*(9.8)*(.7m) + 1/2(.000228)(.82/.03)^2 = 1/2(.83Kg * Vf^2) + 1/2(.43Kg * Vf^2) + (.83Kg * 9.8 * .7m) + 1/2(.000228)(V/.03)^2

-.999761 = .415 Vf^2 + .215 Vf^2 - 2.9498 + .000114 Vf^2

1.95004 = .630114 Vf^2

Vf^2 = 3.09474

Vf = 1.75919 m/s

 

[Doc Al]

1/2(.83Kg * (.82m/s^2)) + 1/2(.43Kg * (.82m/s^2)) - (.25)*(.83Kg)*(9.8)*(.7m) + 1/2(.000228)(.82/.03)^2 = 1/2(.83Kg * Vf^2) + 1/2(.43Kg * Vf^2) + (.83Kg * 9.8 * .7m) + 1/2(.000228)(V/.03)^2

-.999761 = .415 Vf^2 + .215 Vf^2 - 2.9498 + .000114 Vf^2

Double check that first number; I get a significantly different number.

 

[pcandrepair]

I recalculated the -.999761 and got -.914667. Then I got 1.79716 m/s and it's still wrong.

 

[Doc Al]

I recalculated the -.999761 and got -.914667. Then I got 1.79716 m/s and it's still wrong.

See below:

1/2(.83Kg * (.82m/s^2)) + 1/2(.43Kg * (.82m/s^2)) - (.25)*(.83Kg)*(9.8)*(.7m) + 1/2(.000228)(.82/.03)^2 = 1/2(.83Kg * Vf^2) + 1/2(.43Kg * Vf^2) + (.83Kg * 9.8 * .7m) + 1/2(.000228)(V/.03)^2

-.999761 = .415 Vf^2 + .215 Vf^2 - 2.9498 + .000114 Vf^2

Double check the highlighted number (work done by friction). [Edit: That's the change in gravitational PE, not work done by friction.]

 

[pcandrepair]

that number is the M*g*h for the hanging mass which should be right. did I forget to include friction on the right side of the equation?

 

[Doc Al]

that number is the M*g*h for the hanging mass which should be right.

I mislabeled it as work done by friction, but I meant the M*g*h for the hanging mass. Double check that number.

Ah, you have a typo: You wrote .83 instead of .43.

 

[pcandrepair]

(.43Kg)*(9.8m/s^2)*(-.7m) = -2.9498
I'm getting the same number from before.

 

[Doc Al]

one more

1/2(.83Kg * (.82m/s^2)) + 1/2(.43Kg * (.82m/s^2)) - (.25)*(.83Kg)*(9.8)*(.7m) + 1/2(.000228)(.82/.03)^2 = 1/2(.83Kg * Vf^2) + 1/2(.43Kg * Vf^2) + (.83Kg * 9.8 * .7m) + 1/2(.000228)(V/.03)^2

-.999761 = .415 Vf^2 + .215 Vf^2 - 2.9498 + .000114 Vf^2

I think you forgot the (1/.03)^2 factor when you calculated that last term.

 

[pcandrepair]

Now I'm getting 1.60535 m/s.

 

[pcandrepair]

Ok. That 1.60535 is correct. For part b now, would i just use V = Rw to find the angular speed. So,

w = 1.60535 / sqrt(.02^2 + .03^2)
w = 44.5238 and this would be in rad/sec?

 

[pcandrepair]

I used the wrong radius in the previous post. I used .03 and got the right answer. Thanks for all your help Doc Al!