H2 physics question
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1) A cyclist attempts to travel around a vertical circular track of radius 3.0m. Calculate the minimum speed he must have on entry in order to remain in contact with the sphere when he is upside down ( i dont understand what this question wants? )
2) A 20 cm long spring stretches to a length of 22cm when pulled horizontally with a force of 100N. It is placed vertically on the ground and 10.2 kg block is held 10cm above the spring. The block is dropped, hits the spring and compresses it. What is the height of the spring at the point of maximum compression?
Any help please?
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1)Originally posted by lavastar:1) A cyclist attempts to travel around a vertical circular track of radius 3.0m. Calculate the minimum speed he must have on entry in order to remain in contact with the sphere when he is upside down ( i dont understand what this question wants? )
Need to consider force calculations
When you go in circular motion, there's a centripetal acceleration of mv^2/r required. When cyclist is at the very top, the only forces providing this acceleration is his weight and any reaction force from the track.
To just remain in contact, reaction force = 0
Hence, mv^2/r = mg
and v = sqrt(gr) at the top
Supposed he did not add more power when he is on that circular track, we need to take into account KE loss due to increase in PE
so loss in KE = 1/2mV^2 - 1/2mv^2 = mg(6), V = entry speed
V^2 = v^2 + 12g = 15g
V = sqrt(15g) m/s
I think.... -
F = ke, F = 100N, e = 0.02mOriginally posted by lavastar:2) A 20 cm long spring stretches to a length of 22cm when pulled horizontally with a force of 100N. It is placed vertically on the ground and 10.2 kg block is held 10cm above the spring. The block is dropped, hits the spring and compresses it. What is the height of the spring at the point of maximum compression?
k = F/e = 5000N/m
compression gives stored energy
stored energy = 1/2 k e^2
loss in PE = mg(0.1 + e) = gained in stored energy = 1/2 k e^2
100(0.1 + e) = 2500 e^2
1 + 10e = 250 e^2
using formula,
e = 0.0863 or -0.0463 (reject)
compression of 8.63 cm
Height of spring = 20 - 8.63 = 11.4 cm.... Not sure correct a not... -
Its correctOriginally posted by eagle:F = ke, F = 100N, e = 0.02m
k = F/e = 5000N/m
compression gives stored energy
stored energy = 1/2 k e^2
loss in PE = mg(0.1 + e) = gained in stored energy = 1/2 k e^2
100(0.1 + e) = 2500 e^2
1 + 10e = 250 e^2
using formula,
e = 0.0863 or -0.0463 (reject)
compression of 8.63 cm
Height of spring = 20 - 8.63 = 11.4 cm.... Not sure correct a not...
thanks alot -
dis one circular motion and spring constant right?Originally posted by lavastar:1) A cyclist attempts to travel around a vertical circular track of radius 3.0m. Calculate the minimum speed he must have on entry in order to remain in contact with the sphere when he is upside down ( i dont understand what this question wants? )
2) A 20 cm long spring stretches to a length of 22cm when pulled horizontally with a force of 100N. It is placed vertically on the ground and 10.2 kg block is held 10cm above the spring. The block is dropped, hits the spring and compresses it. What is the height of the spring at the point of maximum compression?
Any help please? -
the 1st qn is circular motion 2nd is spring constant + conservation of energy..Originally posted by unclebutcher:dis one circular motion and spring constant right? -
i have one more question.
A 2.20 kg mass, m, on a frictionless table is moving in a circle with radius 0.420 m at a constant speed. m is attached to a 6.00 kg mass, M, by a cord through a hole in the table. Find the speed with which m must move for M to stay at rest.
The solution is to equate the centripetal force of the 2.2kg mass to the weight of the 6 kg mass. but i dun understand why.. can some kind soul explain to me? thanks -
Draw your free body diagram of m.
You will find that when m is moving in a circle, the cord exerts a certain tension force on block m equivalent to the weight of M.
Ok let me illustrate this:
Imagine the cord connecting M and m. When you draw free body diagrams, you will realise that there are 4 forces acting on the cord-blocks system.
F1: Force cord exerts on M
F2: Force M exerts on cord
F3: Force cord exerts on m
F4: Force m exerts on cord
By Newton's 3rd Law, F1 & F2 and F3 & F4 are action-reaction pairs.
As you know, action-reaction pairs are equal in magnitude. When the cord is massless or experiences no acceleration, F1 = F4 and F2 = F3.
In this case, F2 = force M exerts on cord = Weight of M = F3 = force cord exerts on m = centripetal force.