A car goes round a circular track of radius 30.0 m which is banked at an angle of 15o to the horizontal to make it safer when there is ice on the road. What is the maximum safe speed of the car if there is no friction between the tyres and the road?
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please help with question on forces and motion (1 Viewer)
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BlackJack
Vertigo!
For this question you want to balance the gravitys effect on the car against its centripetal motion.
For the car to go around in circles, it must satisfy this equation:
a = v<sup>2</sup>/r
a is the acceleration towards the centre, this is provided by gravity.
Do a little diagram, check the component of gravity along the surface. If I'm not mistaken it is g.sin(15).
.'. v<sup>2</sup> = ra
v = sqrt(30gsin(15))
That should be correct.
...I wonder if I'm missing something... (it's been a while)
EDIT: 15. degrees
For the car to go around in circles, it must satisfy this equation:
a = v<sup>2</sup>/r
a is the acceleration towards the centre, this is provided by gravity.
Do a little diagram, check the component of gravity along the surface. If I'm not mistaken it is g.sin(15).
.'. v<sup>2</sup> = ra
v = sqrt(30gsin(15))
That should be correct.
...I wonder if I'm missing something... (it's been a while)
EDIT: 15. degrees
Last edited:
Ok, this is way more mathematical than our physics course, but I'll give it a shot and pretend that I've done any of mechanics from 4u...
The circular track is banked at 15° so if you construct some triangles and use trigonometry you can work out the force components working up and down the banked road.
Component of gravity working down the hill:
sin(15° ) = F<sub>downhill</sub>/F<sub>g</sub>
F<sub>downhill</sub>= F<sub>g</sub>sin(15° )
Component of force acting up the hill (This is kinda weird without friction, but I'm going to assume centripetal force will come from pushing against the gravitational force...):
cos(15° )= F<sub>c</sub>/F<sub>uphill</sub>
F<sub>uphill</sub> = F<sub>c</sub>/cos(15° )
So the forces are balanced when F<sub>uphill</sub>= F<sub>downhill</sub>:
F<sub>c</sub>/cos(15° )= F<sub>g</sub>sin(15° )
(mass cancels out cause F<sub>c</sub>=mv<sup>2</sup>/r and F<sub>g</sub> = mg)
v<sup>2</sup>/rcos(15° )= gsin(15° )
v = √[rgsin(15° )cos(15° )] I assume that this is the safe max speed 'cause otherwise the forces are unbalanced and the car will slide up the banked curve.
calculating that....
v= √[30x9.8xsin15xcos15]
= √[73.5]
= 8.57 ms<sup>-1</sup> which, to me, seems a completely unlikely answer... I hope my fruitless attempt has helped you
.
The circular track is banked at 15° so if you construct some triangles and use trigonometry you can work out the force components working up and down the banked road.
Component of gravity working down the hill:
sin(15° ) = F<sub>downhill</sub>/F<sub>g</sub>
F<sub>downhill</sub>= F<sub>g</sub>sin(15° )
Component of force acting up the hill (This is kinda weird without friction, but I'm going to assume centripetal force will come from pushing against the gravitational force...):
cos(15° )= F<sub>c</sub>/F<sub>uphill</sub>
F<sub>uphill</sub> = F<sub>c</sub>/cos(15° )
So the forces are balanced when F<sub>uphill</sub>= F<sub>downhill</sub>:
F<sub>c</sub>/cos(15° )= F<sub>g</sub>sin(15° )
(mass cancels out cause F<sub>c</sub>=mv<sup>2</sup>/r and F<sub>g</sub> = mg)
v<sup>2</sup>/rcos(15° )= gsin(15° )
v = √[rgsin(15° )cos(15° )] I assume that this is the safe max speed 'cause otherwise the forces are unbalanced and the car will slide up the banked curve.
calculating that....
v= √[30x9.8xsin15xcos15]
= √[73.5]
= 8.57 ms<sup>-1</sup> which, to me, seems a completely unlikely answer... I hope my fruitless attempt has helped you
.
thunderdax
I AM JESUS LOL!
Using mechanics (4U maths):
Resolving forces horizontally,
Nsin15<sup>o</sup>=(mv<sup>2</sup>)/30
Resolving forces vertically,
Ncos15<sup>o</sup>=mg
dividing,
tan15<sup>o</sup>=v<sup>2</sup>/30g
v=sqrt(30*9.8tan15)
v=8.9ms<sup>-1</sup>
definately not a physics question
Resolving forces horizontally,
Nsin15<sup>o</sup>=(mv<sup>2</sup>)/30
Resolving forces vertically,
Ncos15<sup>o</sup>=mg
dividing,
tan15<sup>o</sup>=v<sup>2</sup>/30g
v=sqrt(30*9.8tan15)
v=8.9ms<sup>-1</sup>
definately not a physics question