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November 01, 2025, 08:48:12 am
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Author Topic: TyErd's Questions pleaseee! :)  (Read 3447 times)  Share 

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/0

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Re: TyErd's Questions pleaseee! :)
« Reply #30 on: May 22, 2010, 07:00:40 pm »
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(Please check this with the answers, it's been a while since I've done mechanics)

If the driver is going to lose contact with the curve at any place, it's going to be at the top of the curve.
At the top of the curve, if we designate DOWN to be positive, then (since both forces point down).
And since the net force when moving in a circle is the same as the centripetal force, this is the same as:



The minimum possible normal force you need to stay in contact with the road is . So we want to have

at the top of the circle.

Hence, at the top of the circle.

But the question asks for the speed you need at the bottom of the circle. To find this we can use conservation of energy:





(We assume that the driver takes their foot off the pedal once they reach the curve, so there is no extra energy from the engine)

« Last Edit: May 22, 2010, 07:03:09 pm by /0 »
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TyErd

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Re: TyErd's Questions pleaseee! :)
« Reply #31 on: May 22, 2010, 07:54:30 pm »
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answer should be 26.2m/s
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/0

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Re: TyErd's Questions pleaseee! :)
« Reply #32 on: May 22, 2010, 08:10:55 pm »
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Oh, perhaps it was asking for the speed at the top, which is just

The question is a bit ambiguous, and it should be asking for the 'minimum' speed, not the 'maximum' speed.
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TyErd

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Re: TyErd's Questions pleaseee! :)
« Reply #33 on: May 23, 2010, 10:40:21 am »
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im really strugling to understand these type of questions. stuff like "force exerted on the road by the tyres" or "force exerted by the tyres on the road" really confuses me. help please :)
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appianway

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Re: TyErd's Questions pleaseee! :)
« Reply #34 on: May 23, 2010, 11:54:04 am »
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The diagram's tiny..

... but think about this. If the car is undergoing circular motion, what has to be the net force? Obviously the road's the only thing that can provide that force, and this force acts on the tyre. Due to newton's third law, the tyre has to exert an equal and opposite force on the road.
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TyErd

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Re: TyErd's Questions pleaseee! :)
« Reply #35 on: May 23, 2010, 01:03:56 pm »
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click the diagram to make it bigger.
so does that mean, the road therefore exerts a force on the tyres towards the centre of the circle? which means the tyres exert a force opposite in direction equal in magitude?
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appianway

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Re: TyErd's Questions pleaseee! :)
« Reply #36 on: May 23, 2010, 02:19:21 pm »
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Exactly :)
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TyErd

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Re: TyErd's Questions pleaseee! :)
« Reply #37 on: May 23, 2010, 05:09:45 pm »
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The acceleration of Tom is zero at some distance below the bridge. Calculate that distance. The cord has a spring constant of 254 N/m

not sure about this one
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TyErd

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Re: TyErd's Questions pleaseee! :)
« Reply #38 on: May 23, 2010, 05:32:46 pm »
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Find the change in momentum (magnitude and direction) of Tom and the motorbike from the
moment of taking off to the moment just before landing.

This one has me confused as well
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"Don’t ever let somebody tell you you can’t do something, not even me.  Alright?  You got a dream, you gotta protect it.  People can’t do something themselves, they wanna tell you you can’t do it.  If you want something, go get it, period." - Chris Gardner
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