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I know how to find the horizontal acceleration but I don't know where to go from there.

Any hints?

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- Thread starter zekester
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- #1

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I know how to find the horizontal acceleration but I don't know where to go from there.

Any hints?

- #2

Doc Al

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Check out this thread: https://www.physicsforums.com/showthread.php?s=&threadid=10101Originally posted by zekester

I know how to find the horizontal acceleration but I don't know where to go from there.

- #3

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There's an hard way and an easy way.

The easy way is to use conservation of angular momentum.

JMD

The easy way is to use conservation of angular momentum.

JMD

- #4

Doc Al

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Angular momentum is not conserved. (The friction exerts a torque.)Originally posted by nbo10

The easy way is to use conservation of angular momentum.

- #5

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Angular momentum is conserved about a point, you have to find the correct point.

and if your looking for the FINAL speed, it's going to be zero.

JMD

and if your looking for the FINAL speed, it's going to be zero.

JMD

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- #6

HallsofIvy

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Angular momentum is conserved about a point, you have to find the correct point.

and if your looking for the FINAL speed, it's going to be zero.

Aren't those two statements contradictory? If the final speed is 0, then the angular momentum is 0 which is not true while it's rolling.

Actually once it gets to pure roling without sliding, the friction is no longer a factor (that's the whole point of the wheel!).

- #7

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alpha = (5 * coeff friction * g)/2R

omega = (5 * coeff friction * g * t)/2R

the 8 m/s is beginning translation motion right? but i cant figure out where to begin. ive gone through a few sheets of paper, and i dont think he gave us enough to start with.

- #8

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okay, using 8 = R*omega, i was able to find a time. i did 8 = R (5 * coeff friction *g * t)/2r. i ended up with a time of 5.4 seconds. but im stuck now.

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- #9

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alpha = (5 * mu * g) 2*R

alpha = a/r, so a/r = (5 * mu * g)/ 2*R. with this i ended up with an a of 1.47.

i than used this 1.47 in omega = initial omega(which is zero because it was not rotating) + alpha * time.

i found time using the formula:

t = 2 * initial velocity/7 * mu * g, which equals 3.89 s.

than i went with v/r = 0 + 1.47 * t/R

the r's cancel and i ended up with a final velocity of 5.7 m/s.

have i done everything correct?

- #10

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correct.. You can also use conservation of Angular momentum, about the point where the ball makes contact with the floor.

When the ball first starts

[itex]l=mrv_0

[/itex]

when the ball starts to roll without sliping

[itex]l=I \omega [/itex]

[itex]I = \frac{2}{5}mr^2 + mr^2 [/itex]

for rolling without slipping you can find

[itex]\omega = \frac{v}{r} [/itex]

and you can solve these and find

[itex]v = \frac{5}{7}v_0

[/itex]

Which is 5.7 m/s

I might have to edit this to get the LaTex to work.

JMD

When the ball first starts

[itex]l=mrv_0

[/itex]

when the ball starts to roll without sliping

[itex]l=I \omega [/itex]

[itex]I = \frac{2}{5}mr^2 + mr^2 [/itex]

for rolling without slipping you can find

[itex]\omega = \frac{v}{r} [/itex]

and you can solve these and find

[itex]v = \frac{5}{7}v_0

[/itex]

Which is 5.7 m/s

I might have to edit this to get the LaTex to work.

JMD

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- #11

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Originally posted by HallsofIvy

Aren't those two statements contradictory? If the final speed is 0, then the angular momentum is 0 which is not true while it's rolling.

They are contradictory. I've been ill with the flu and still not thinking clearly all the time. The final speed won't be zero.

JMD

- #12

Doc Al

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Excellent. Hadn't thought of that.Originally posted by nbo10

correct.. You can also use conservation of Angular momentum, about the point where the ball makes contact with the floor.

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