Nonuniform circular motion toy train

In summary, the problem involves a toy train rolling around a horizontal track with a diameter of 1.0m and a coefficient of rolling friction of 0.10. The task is to determine the train's angular acceleration after being released and the time it takes for the train to stop if it was initially released with an angular speed of 30rpm. To solve the problem, we can use the equations At = r(alpha) and A = v^2/r, as well as Newton's second law or the work-energy theorem to account for the friction coefficient. The tangential acceleration can be found by taking the derivative of the angular velocity with respect to time.
  • #1
Jennifer001
22
0

Homework Statement


a toy train rolls around a horizontal 1.0m diameter track. the coeffifient of rolling friction is 0.10

a. what is the magnitude of the train's angular acceleration after it is released
b. how long does it take the train to stop if it's released with an angular speed of 30rpm.



Homework Equations



At=r(alpha)
A=v^2/r
Ff=uk*ma=uk*mg


The Attempt at a Solution



i know I am suppose to look for the angular acceleration but i don't know what the tangential acceleration is, is there a relationship between centripetal acceleration and tangential .. also i don't know how to include the friction coieff. into an equation..

PLEASE HELP IM SO LOST!



so i have
 
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  • #2
Jennifer001 said:
a toy train rolls around a horizontal 1.0m diameter track. the coeffifient of rolling friction is 0.10

a. what is the magnitude of the train's angular acceleration after it is released
b. how long does it take the train to stop if it's released with an angular speed of 30rpm.

i know I am suppose to look for the angular acceleration but i don't know what the tangential acceleration is, is there a relationship between centripetal acceleration and tangential .. also i don't know how to include the friction coieff. into an equation..

Hi Jennifer001! :smile:

For a constant radius, the tangential acceleration is just dv/dt.

And you put the friction into Newton's second law to find the acceleration (or you could just use the work-energy theorem ), exactly as on a straight track.
 
  • #3
to find the tangential acceleration first, which is equal to the centripetal acceleration, since the train is moving in a circular motion. The formula for centripetal acceleration is A=v^2/r, where v is the tangential velocity and r is the radius of the circle. Since the train is rolling at a constant speed, its tangential velocity is equal to its angular velocity multiplied by the radius. So, v=ωr. Substituting this into the equation, we get A=ω^2r.

Next, we need to find the angular acceleration, which is equal to the tangential acceleration divided by the radius. So, α=A/r. Substituting the value of A we found earlier, we get α=ω^2.

Now, to find the magnitude of the angular acceleration, we need to find the value of ω. We know that the train is released with an angular speed of 30rpm (revolutions per minute). We can convert this to radians per second by multiplying it by 2π/60. So, ω=30*2π/60=π/3 rad/s.

Substituting this value into our equation for α, we get α=(π/3)^2=π^2/9 rad/s^2. This is the magnitude of the train's angular acceleration.

For part b, we can use the formula ω=ω0+αt, where ω0 is the initial angular velocity, α is the angular acceleration, and t is the time. We are given the initial angular velocity (ω0) as 30rpm. We can convert this to radians per second by multiplying it by 2π/60. So, ω0=30*2π/60=π/3 rad/s.

We are also given the angular acceleration (α) as we found in part a. So, substituting these values into the equation, we get 0=π/3+π^2/9*t. Solving for t, we get t=9/π seconds. This is the time it takes for the train to stop.
 

What is nonuniform circular motion?

Nonuniform circular motion is the motion of an object along a circular path with varying speed. This means that the object is constantly changing its speed as it moves along the circular path.

How does a toy train demonstrate nonuniform circular motion?

A toy train can demonstrate nonuniform circular motion by moving along a circular track at varying speeds. The train may slow down or speed up as it moves around the track, showing the changing speed characteristic of nonuniform circular motion.

What causes nonuniform circular motion in a toy train?

Nonuniform circular motion in a toy train is caused by the force of gravity acting on the train as it moves along the circular track. This force, along with other factors such as friction, can cause the train to speed up or slow down at different points along the track.

How can nonuniform circular motion in a toy train be calculated?

The acceleration of a toy train moving along a circular track can be calculated using the formula a = v^2/r, where a is the acceleration, v is the speed, and r is the radius of the circular track. This formula can be used to determine the train's acceleration at any point along the track.

What real-life applications use nonuniform circular motion?

Nonuniform circular motion has many real-life applications, such as in amusement park rides, car racing, and satellite orbiting. It is also used in the study of planetary motion and the movement of celestial bodies in the universe.

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