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17 Motion in a circle (17.1 Uniform circular motion (Angular displacement…
17 Motion in a circle
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17.2 On the road
Over the top of a hill
Support force is directly upwards from the road opposite the weight. Resultant force is difference between weight and support force. Difference acts towards centre of curvature of hill as centripetal force
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If vehicle loses contact, speed is equal or greter than particular speed. So S=0 -> mg=mv2/r
On a roundabout
Centripetal force is provided by sideways force of friction between the vehicles tyres and road surface, force of friction F = mv2/r
For no skidding or slipping, friction must be less than limiting value. Limiting force of friction f0 = mv2/r = μmg
On a banked track
Without banking, centripetal force on road vehicle is provided only by sideways friction
On banked track, speed can be higher. No sideways friction means N1 and N2 coming from both sides of vehicle acting perpendicular to bank must act as centripetal force
Condition for no sideways friction tanθ = v2/gr so there is no sideways friction if the speed v is v2 = grtanθ
At the fairground
The Big Dipper
Pushes you into your seat as you pass through dip. Difference between support force and weight acts as centripetal force
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The very long swing
Max speed = lowest point. Can be worked out from equating gain of KE to loss of PD: 1/2xmv2=mgh therefore, v2=2gh
At lowest point S from rope is opposite direction to weight. S-mg acts as centre of circular path so centripetal force, S-mg = mv2/L = 2mgh/L
The Big Wheel
At max height, reaction force is downwards so resultant force at this position is equal to mg+R. Reaction force and weight provides centripetal force. Therefor mg+R = mv2/r
At particular speed such that v=gr, then r=0 so there would be no force on the person due to the wheel