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November 08, 2025, 03:57:00 am

Author Topic: Circular Motion Question  (Read 1321 times)  Share 

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kevinn

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Circular Motion Question
« on: May 06, 2010, 06:31:59 pm »
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A cyclist rounds a bend of radius 10m with a speed of 9m/s.  The surfaces of the road is horizontal.  The cyclist is forced to lean at an angle of 20 degrees to the vertical to just take the bend successfully.  The total sideways frictional force on the tyres is 360N.  The cycle has a mass of 20kg.  What is the mass of the cyclist?

Any help would be appreciated :)

appianway

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Re: Circular Motion Question
« Reply #1 on: May 06, 2010, 10:30:46 pm »
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The total sum of the forces on the horizontal axis will give you the centripetal force. This comes from the frictional force (which is given), and a component of the normal force in this direction. As the normal force is mg, you can find out what this component has to be and hence the value of m.

median

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Re: Circular Motion Question
« Reply #2 on: May 31, 2010, 09:18:19 pm »
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Can someone work me through the working

Heres the track Im going along.

total centripetal acceleration=v^2/r=8.1ms^-2
horizontal normal force contributing to centripetal force=8.1*mass-360
Total normal reaction force of cyclist=9.8*mass=(centripetal force)/sin(20)
I cant figure out the normal reaction force though. Is the fact that a 20 degree angle is required to just make the bend without slipping relevant??




barydos

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Re: Circular Motion Question
« Reply #3 on: February 03, 2013, 06:22:56 pm »
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Sorry for this huge revival of this thread, but I'd like to know how to answer this question.
I can't seem to get my head around it. Is the sideways frictional force the centripetal force? :S
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paulsterio

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Re: Circular Motion Question
« Reply #4 on: February 03, 2013, 06:32:25 pm »
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Sorry for this huge revival of this thread, but I'd like to know how to answer this question.
I can't seem to get my head around it. Is the sideways frictional force the centripetal force? :S

Yes, it's the only force acting in that direction, so the centripetal force (net force - sum of all forces) is equal to the friction. :)

barydos

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Re: Circular Motion Question
« Reply #5 on: February 03, 2013, 07:31:26 pm »
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Great, got it thanks :)
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Jaswinder

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Re: Circular Motion Question
« Reply #6 on: February 03, 2013, 07:45:27 pm »
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Im just curious, what answer did you get?  :-\

barydos

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Re: Circular Motion Question
« Reply #7 on: February 04, 2013, 05:26:38 pm »
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Im just curious, what answer did you get?  :-\

Assuming gravity is 9.8m/s^2, I found the mass of the cyclist to be ~80.93kg.
If I answered wrongly, please feel free to correct me!
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paulsterio

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Re: Circular Motion Question
« Reply #8 on: February 04, 2013, 06:02:27 pm »
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Use g = 10 N/kg for physics please, 9.8 is more accurate, but that's for specialist mathematics.

barydos

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Re: Circular Motion Question
« Reply #9 on: February 04, 2013, 06:14:38 pm »
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In that case I found the mass of the cyclist to be 78.9 kg. :)
2012: Methods [47] | Chinese SL [35]
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ATAR: 99.55