Circular Motion and tangential acceleration Problem

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A car accelerates on a circular track with a radius of 100m and a tangential acceleration of 0.1m/s², starting from rest. After 10 seconds, it travels 5m and reaches a velocity of 1m/s. The frictional force acting on the car is calculated using the formula (mv²)/r, yielding a value of 0.142 Newton. However, it is noted that this only accounts for one component of the frictional force, as the tangential acceleration is also due to friction. To find the total frictional force, both radial and tangential accelerations must be combined vectorially.
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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\sqrt{2}.


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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Start by identifying the forces on the car. Also, what is the acceleration of the car?
 
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\sqrt{2}
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\sqrt{2}
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?
 
Yes, but remember that the components are vectors, so you need to add them vectorially to get the total acceleration.
 
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