Circular Motion and tangential acceleration Problem

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animesh27194
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A car starts moving on a circular track of radius 100m from rest at t=0. The tangential acceleration of the car is 0.1m/s2
Then find out the frictional force acting on the car at t=10s. It is given that the car is not skidding during this period and the mass of the car is 10[tex]\sqrt{2}[/tex].


I got that the distance traveled after 10 seconds as 5m and the velocity of the car at the end of 10 seconds as 1m/s

But how do i proceed after that? :confused:
 
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I treid solving the problem, but I don't know if it is correct...

Using the given tangential acceleration, i found out the distance after 10 seconds and thereby, its velocity.

s = ut+(1/2)at2
s = (1/2)(0.1)(100)
s= 5m

Also, v2 = u2 + 2as
v2 = 2(0.1)(5)
Hence, velocity at t=10s is 1m/s

Using this value of v in the formula frictional force = (mv2)/r
Here, m=10[tex]\sqrt{2}[/tex]
v=1 m/s
r=100m

Solving this, I get fictional force as 0.142 Newton.

Is this right?:redface:
 
animesh27194 said:
I treid solving the problem, but I don't know if it is correct...

Using the given tangential acceleration, i found out the distance after 10 seconds and thereby, its velocity.

s = ut+(1/2)at2
s = (1/2)(0.1)(100)
s= 5m

Also, v2 = u2 + 2as
v2 = 2(0.1)(5)
Hence, velocity at t=10s is 1m/s
This is correct, but you could have used v=u+at to find the speed of the car a little more directly.
Using this value of v in the formula frictional force = (mv2)/r
Here, m=10[tex]\sqrt{2}[/tex]
v=1 m/s
r=100m

Solving this, I get fictional force as 0.142 Newton.

Is this right?:redface:
Almost. You've found one component of the frictional force. The tangential acceleration of the car is also the result of friction between the tires and the road.
 
So, should i add the radial and the tangential acceleration to get the total acceleration?
And I multiply that with the mass of the car to get the frictional force?