Centripetal force and acceleration problem

[ScienceGirl90]

Homework Statement

A 2500 kg car starts from rest and travels along a circular track with a radius of 55.0 meters. The car starts out at a position due East of the center of the track, heading due North initially around the track, which is counterclockwise. The car increases its speed by 105 m/s every minute around the track. How fast is the car traveling when it is 1/4 of the distance around the track? State the car's tangental velocity.

Homework Equations

v=2*pi*r/period

The Attempt at a Solution

I am unsure of how to begin this problem when I'm looking for only 1/4 of the distance. I'm also having a hard time finding the period of one full rotation around the track. Any help would be appreciated! Thanks.

 

[Delphi51]

In this case you are better off doing the magnitude and the direction separately. Figure out the distance for the quarter circle and translate the "105 m/s every minute" into an acceleration in m/s. Then treat it like a linear motion problem and just use your constant acceleration formulas to find the final speed.

 

[ScienceGirl90]

Ok, thank you. Would you mind explaining how I get 105 m/s every minute to m/s for acceleration?

 

[venkatg]

105m/s per minute translates to an acceleration a = of 105/60 = 1.75 m/sec/sec

To cover 1/4 th the circumference of a circle of radius is (1/4) * 2 * pi * r

this gives the distance "S"

Now use the formula velocity after covering S is =

V*V = 2* a * S

 

[rwisz]

I would approach this problem by first identifying the key variables and equations that relate to centripetal force and acceleration. In this case, the key variables are the mass of the car (2500 kg), the radius of the track (55.0 meters), and the acceleration (105 m/s) that the car experiences every minute. The equation that relates these variables is F=ma, where F is the centripetal force, m is the mass, and a is the acceleration.

To find the speed of the car when it is 1/4 of the distance around the track, we can use the equation for tangential velocity: v=2πr/period. However, as you mentioned, we need to know the period of one full rotation around the track. To find this, we can use the equation for centripetal acceleration: a=v^2/r. Rearranging this equation, we get v=√(ar). Plugging in the given values, we get v=√(105*55)= 40.81 m/s as the tangential velocity.

Now, to find the period, we can use the equation T=2πr/v (where T is the period and v is the tangential velocity). Plugging in the values, we get T=2π*55/40.81= 3.40 seconds as the period of one full rotation around the track.

Since we are looking for the speed of the car when it is 1/4 of the distance around the track, we can divide the period by 4 to get the time it takes for the car to travel 1/4 of the distance around the track. This is equal to 3.40/4= 0.85 seconds.

Now, we can use the equation v=2πr/T to find the speed of the car when it is 1/4 of the distance around the track. Plugging in the values, we get v=2π*55/0.85= 409.76 m/s as the speed of the car when it is 1/4 of the distance around the track.

In conclusion, the car is traveling at a speed of 409.76 m/s when it is 1/4 of the distance around the track. Its tangential velocity is 40.81 m/s.