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Simple mechanics/physics circular motion problem?

If you have a vehicle moving on a banked corner via uniform circular motion. The centripetal force acts horizontally towards the centre of the circle. Why doesn't the centripetal force act down the inclined plane instead?

Reply 1

Nuclear Ghost

Original post by bobbricks

You have to consider the component of force which acts directly horizontally, which includes weight and friction, and not along the inclined plane.

Reply 2

bobbricks

Also, I'm stuck on part C here:
https://isaacphysics.org/questions/roundabout_num?board=d84c1697-c41f-47eb-8ce3-4d0e3d723972

Fr=μR
I resolved perpendicular to the slope so R=mgcos30
so Fr=μmgcos30
I resolved horizontally so:
mgsin30cos30-μmg(cos30)(cos30)=mv^2/r

The m's cancel out and I
Rearranged it and ended up with v=3.2ms^-1

But that isn't correct- why?

EDIT: I just watched the video in the hint which indicates that the Friction acts down the slope but I'm not sure why since I thought that the friction acts in the direction that opposes motion so it would act up the slope (as its motion is inwards) ?

(edited 9 years ago)

Reply 3

atsruser

Original post by bobbricks

If a particle is travelling with constant speed in a circle, the net force on the particle must be

a) of constant strength
b) always pointing towards the centre of the circle

This follows from Newton II and from the analysis of the direction of acceleration of a particle moving in a circle at constant speed.

So in your example, since the vehicle is moving in a horizontal circle, the force maintaining that motion must act horizontally too. All of the forces on the car must add up vectorially to produce this centripetal force.