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Rotational Motion Homework

  1. Feb 18, 2004 #1
    I was just wondering if anyone has a few minutes if they could check my answers for my homework. I have 7 questions on Rotaional Motion

    1)A potter's wheel moves from rest to an angular speed of .20rev/s in 30s. Assuming constant angular acceleration, what is its angular acceleration in rads/s^2

    á=.0419 rads/s^2

    2)A tire placed on a balencing machine in a service station starts from rest and turns through 4.7 revs in 1.2s before reaching its final angular speed. Assuming that the angular acceleration is constant calculate á.

    á=20.497 rads/s^2

    3)The Emerald Suite, a revolving resturant at the top of the Space Needle in Seatle, makes a complete turn once every hour. What is the tangential speed of a customer sitting 12.5 m from the resturants center?

    Vt=.0218 m/s

    4)What is the magnitude of centripetal acceleration of the customer in problem 3

    Ac=3.803*10^-5 m/s^2

    5)A 515kg Roller Coster rolls down past point A and then up past point B. Using the following diagram answer a and b below.
    [​IMG]
    a)If the vehicle has a speed of 20 m/s at point A, what is the force of the track on the vehicle at this point?
    b)What is the maximum speed the vehicle can have at point B for g to hold it on track?

    a)F=1030N
    b)Vt<=12.165m/s

    6)A 12500N car traveling at 50km/h rounds a curve of radius 200m find the following
    a)the centripetal acceleration of the car
    b)the force that maintains centripetal acceleration
    c)the minimum coeffiecent of kinetic friction between the tires and the road so that the car can safely round the curve.

    a)Ac=12.5 m/s^2
    b)Fc=17201.8 N
    c)Coefficient>=1.274

    7)A copper Block rests 30cm from the center of a steel turntable. The coefficient of static friction between the block and the turntable is .53. The turntable starts from rest and rotates with a constant angular acceleration of .5 rads/s^2. After what time interval will the block start top slip on the turntable.

    Ät=8.3265s

    Any help will be greatly appreciated.
     
  2. jcsd
  3. Feb 19, 2004 #2

    jamesrc

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    Science Advisor
    Gold Member

    1. looks good

    2. double check your work (it looks like you missed a factor of 2 in there)

    3. looks good

    4. looks good

    5. I'm not sure what you did here.
    a. in this position, the normal force should be equal to the sum of the weight and the centripetal force.

    b. this one looks about right, though I'm not sure where your last few digits came from (to be specific, I get ~12.13 m/s). If you set the weight equal to the centripetal force, then you did it correctly, just run the numbers one more time.

    6. Watch your units and try this one again. This problem is a straightforward application of the v2/r formula (then multiply by mass to get force, then divide the second result by weight to get coefficient).

    7. Looks good (you don't need to report that many digits, though)
     
  4. Feb 19, 2004 #3
    Thank you very much for your help. I found out what I did on 5a I just solved for Centripetal force and forgot all about the weight of the car. I'll redo the others I got wrong and post back later. Thank you again.
     
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