Circular Motion and tangential acceleration Problem

AI Thread Summary
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.
animesh27194
Messages
6
Reaction score
0
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:
 
Physics news on Phys.org
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.
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Correct statement about a reservoir with an outlet pipe'
The answer to this question is statements (ii) and (iv) are correct. (i) This is FALSE because the speed of water in the tap is greater than speed at the water surface (ii) I don't even understand this statement. What does the "seal" part have to do with water flowing out? Won't the water still flow out through the tap until the tank is empty whether the reservoir is sealed or not? (iii) In my opinion, this statement would be correct. Increasing the gravitational potential energy of the...
Back
Top