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Circular motion problem -- A ball rolling down an incline

  • Thread starter bijou1
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Homework Statement


A 1 kg ball with a radius of 20 cm rolls down a 5 m high inclined plane. Its speed at the bottom is 8 m/s. How many revolutions per second is the ball making when at the bottom of the plane?

Homework Equations


circumference = 2πr
velocity = distance / time = circumference / time
period (T) = time in seconds
radius = 20 cm ⇒ 0.2 m
mass = 1 kg
velocity at bottom of plane = 8 m/s

The Attempt at a Solution


The velocity and circumference of a circle is related through the equation
⇒ velocity = circumference / period
⇒velocity = 2πr / T
⇒velocity = ( 2 * 3.14 * 0.2 m ) / T
⇒8 m/s = 1.3 m / T
⇒T = 0.16 s
8 m /1 s so 8 m / 0.16 s = 50 m/s ??

I have trouble getting the revolutions/second for the ball when it is at the bottom of the plane.
Is revolutions/second the equivalency of meters/second for velocity? I know my approach to this problem is incorrect, and so any help would be great. Thanks.
 

Answers and Replies

  • #2
Bystander
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⇒ velocity = circumference / period
⇒velocity = 2πr / T
⇒velocity = ( 2 * 3.14 * 0.2 m ) / T
⇒8 m/s = 1.3 m / T
You've done fine to this point. Now, just solve for "T." Just as an incidental comment on notation, upper case "T" is usually used for temperature, and time is generally indicated with a lower case "t."
 
  • #3
ehild
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Homework Statement


A 1 kg ball with a radius of 20 cm rolls down a 5 m high inclined plane. Its speed at the bottom is 8 m/s. How many revolutions per second is the ball making when at the bottom of the plane?

Homework Equations


circumference = 2πr
velocity = distance / time = circumference / time
period (T) = time in seconds
radius = 20 cm ⇒ 0.2 m
mass = 1 kg
velocity at bottom of plane = 8 m/s

The Attempt at a Solution


The velocity and circumference of a circle is related through the equation
⇒ velocity = circumference / period
⇒velocity = 2πr / T
⇒velocity = ( 2 * 3.14 * 0.2 m ) / T
⇒8 m/s = 1.3 m / T
⇒T = 0.16 s
8 m /1 s so 8 m / 0.16 s = 50 m/s ??

I have trouble getting the revolutions/second for the ball when it is at the bottom of the plane.
Is revolutions/second the equivalency of meters/second for velocity? I know my approach to this problem is incorrect, and so any help would be great. Thanks.
The ball makes a full revolution in T seconds, so the number of revolutions per second is 1/T.
The angular velocity is ω=2π/T (radian/second) The linear velocity and angular velocity are related as v=ωr.
 
  • #4
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The ball makes a full revolution in T seconds, so the number of revolutions per second is 1/T.
The angular velocity is ω=2π/T (radian/second) The linear velocity and angular velocity are related as v=ωr.
Hi, so according to this equation 1/T, the ball makes an "x" number of revolutions per second. Then, from above I found that T = 0.16 s, therefore
1/0.16 s = 6.3 revolutions / second. How did you get the equation 1/T for the number of revolutions/second? Thanks for your help!
 
  • #5
haruspex
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How did you get the equation 1/T for the number of revolutions/second?
If it makes one revolution in T seconds, how long does it take to make N revolutions? What do you get if you set that time equal to 1 second?
 
  • #6
ehild
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Hi, so according to this equation 1/T, the ball makes an "x" number of revolutions per second. Then, from above I found that T = 0.16 s, therefore
1/0.16 s = 6.3 revolutions / second. How did you get the equation 1/T for the number of revolutions/second? Thanks for your help!
It is the same as with the "common" velocity : You walk to reach to your destiny 200 km away in 2.5 hours, what is the distance your car covers in 1 hour? ( The speed of the car is distance covered / time period)
If the ball rotates by the angle θ in t time , the angular velocity is ω=θ/t . Here, the angle rolls the play of "distance covered".
You can measure the angle in complete turns, that is, in terms of 2π radians. For example, If the ball turns 1000 in 10 seconds, the number of revolutions is 1000/10 = 100 in one second. If the ball needs 0.1 s to turn once, then the number of revolutions is 1/0.1 = 10 in one second. If the ball makes a complete turn in T time, the number of revolutions per second is 1/T .
 
  • #7
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if its rolling without slipping, the surface of the ball is also travelling at 8 m/s, so, how many circumferences fit into 8 metres ?
 
  • #8
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It is the same as with the "common" velocity : You walk to reach to your destiny 200 km away in 2.5 hours, what is the distance your car covers in 1 hour? ( The speed of the car is distance covered / time period)
If the ball rotates by the angle θ in t time , the angular velocity is ω=θ/t . Here, the angle rolls the play of "distance covered".
You can measure the angle in complete turns, that is, in terms of 2π radians. For example, If the ball turns 1000 in 10 seconds, the number of revolutions is 1000/10 = 100 in one second. If the ball needs 0.1 s to turn once, then the number of revolutions is 1/0.1 = 10 in one second. If the ball makes a complete turn in T time, the number of revolutions per second is 1/T .
Hi, so the "distance covered" is 2*3.14*(0.2 m) = 1.3 m and finding "t" through v = 2 π r / t , yielding the answer t = 0.16 s and knowing that the ball makes one revolution in
0.16 s gives the answer (1 revolution / 0.16s ) = 6.3 revolutions / second. Initially I thought I had to divide 1.3 m / 8 m/s, which is completely wrong. Thank you so much for your explanation, and taking the time to explain to me in step by step detail!
 

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