Help?

[Mongaa]

Hey everyone I was just wondering if you guys could help me with a couple questions.  I'll start with this one.

A car approaches a speed bump sign with a speed advisory of 25 km/h.  Assume this speed corresponds to the care 'jumping' over the speed bump, what is the radius of the speed bump?
Thanks

[availn]

Hey everyone I was just wondering if you guys could help me with a couple questions.  I'll start with this one.

A car approaches a speed bump sign with a speed advisory of 25 km/h.  Assume this speed corresponds to the care 'jumping' over the speed bump, what is the radius of the speed bump?
Thanks

I had to google what a speed advisory was, I'm so bad lol. Anyways, the mention of a radius is a hint, that this is a circular motion question. The speed bump is just a large cylinder embedded in the ground. If you go too fast, then you'll lift off the ground. The centripetal acceleration is gravity pulling you down, keeping you on the ground. You'll have to convert 25kmh^-1 to about 6.94ms^-1 In circular motion:

a = v²/r
10ms^-2 = (6.94ms^-1)²/r
r = 48.2m²s^-2/10ms^-2
∴ r = 4.8m

[Dayman]

I'm sorry to intrude but did I  just read that there is such thing as 4.8m radius speed bump?

[Dayman]

Or is it the width wise feel so dumb lol

[Mongaa]

mission_impossible these are probably just hypothetical questions haha, and availn don't worry about part b.) of that question I just realised it doesn't make sense.

[availn]

Thank you so much ! 
Can you help me with this question as well?

A toy car of mass 100 grams enters a loop with diameter 450 mm.

a.) What is the minimum height the car need to be launched down the ramp to make it through the loop (assuming no air resistance) ?
What I did was equate the centripetal acceleration to gravity since that is what it must equal in order to make it through the loop, and so
ac=g
v^2/r=g
v=sqrt(gr)
  = sqrt(10*0.23m)
  =1.52 m/s

And since Kef=Pei

0.5mv^2=mgh
giving me h=0.5v^2/g
so 0.5*(1.52)^2/10
= 0.12m which is the minimum height.
Is this correct?

b.) If the car is dropped from half this height, but is now secured to the track with a magnet, what is the minimum load(in N) the magnet must carry for the car to make it through the loop?

I don't understand this one.
Thanks

It's kinda hard for me to picture this question, especially b) as you said. But one thing you must remember with these sorts of questions is where you're making the calculation from. 0.12m is the minimum height above the top of the loop for it to have a velocity of 1.52ms^-1 at the top of the loop. If the ramp leads to the bottom of the loop (I'm assuming it does), then the minimum starting height up the ramp would be 0.12m + 0.45m = 0.57m.

I'm sorry to intrude but did I  just read that there is such thing as 4.8m radius speed bump?

You're correct, the speed bump has that radius; that really got me too when I first came across this sort of question. But most of the bump is sunk into the ground. This way, only a small portion of the cylinder protrudes above the ground.

[Dayman]

ohhh okay that sounds rreasonable enough but seriously imagine a 4.8m figure potruding off the ground might as well have a wall lol