ATAR Notes: Forum

VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Physics => Topic started by: leflyi on March 10, 2013, 08:24:57 pm

Title: Banked Curves Circular Motion Question
Post by: leflyi on March 10, 2013, 08:24:57 pm

A cycling velodrome has a turn that is banked at 33 degrees to the horizontal. The radius of the track at this point is 28m.

- Determine the speed at which a cyclist of mass 55kg would experience no sideways    force on their bike as they ride this section of track

Just a little unsure of how to tackle this question, I've tried multiple ways but no luck..

Any hints would be appreciated!

thanks- Leflyi

Title: Re: Banked Curves Circular Motion Question
Post by: availn on March 10, 2013, 09:06:10 pm

If the track is frictionless, then the only force keeping the cyclist in centripetal motion is the horizontal component of the cyclist's normal reaction. Draw a diagram for this, and you will find:

N_horizontal = mg sinθ
N_horizontal = 55kg x 10ms^-2 x sin(33°)
N_horizontal = 300N

This is the centripetal force acting on the cyclist. From here it's quite simple:

F = mv² / r
300N = 55kg x v² / 28m
v² = 152Nm kg^-1
v = 12.3 ms^-1

On a side note, the cyclist's weight is not required in this problem, as it ends up cancelling itself out. You could solve this through acceleration only.

mg sinθ = mv² / r
∴ g sinθ = v² / r

Title: Re: Banked Curves Circular Motion Question
Post by: leflyi on March 10, 2013, 09:16:33 pm

Thanks for the reply, yet i figured out what I had been doing incorrectly, I incorrectly added my vectors... :O Hahha

Thanks anyway.

Edit

I solved through this method: