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 Form 4 Physics uniform circular motion online lessons

Motion in a vertical circle with worked examples

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Answer Text:
Motion in a vertical circle:
-Consider a mass ‘m’ tied to a string of length ‘r’ and moving in a vertical circle as shown below.
At position 1– both weight (mg) and tension T are in the same direction and the centripetal force is provided by both, hence T1 + mg = #(mv^2)/r#. T1 = #(mv^2)/r# – mg.(The velocity decreases as T1 decreases since mg is constant).T1 will be zero when #(mv^2)/r# = mg and thus v = #sqrt rg#
- this is the value of minimum speed at position 1 which keeps the body in a circle and at this time when T = 0 the string begins to slacken. At position 2– the ‘mg’ has no component towards the centre thus playing no part in providing the centripetal force but is provided by the string alone.
T2 = #(mv^2)/r#/r
At position 3– ‘mg’ and T are in opposite directions,
therefore;
T3 – mg = #(mv^2)/r#; T3 = #(mv^2)/r# + mg– indicates that the greatest value of tension is at T3 or at the bottom of the circular path.
Examples
1. A ball of mass# 2.5 times 10^-2# kg is tied to a string and whirled in a horizontal circular path at a speed of #5.0 ms^-2#.
- If the string is 2.0 m long, what centripetal force does the string exert on the ball?
Solution
Fc = #(mv^2)/r# =# (2.5 times 10^-2) times 52 /2.0 = 0.31 N.#
2. A car of mass #6.0 times 10^3# kg is driven around a horizontal curve of radius 250 m. if the force of friction between the tyres and the road is 21,000
N. What is the maximum speed that the car can be driven at on a bend without going off the road?
solution:
Fc= force of friction=21,000,
also Fc=#mv^2)/r#, hence
21000=#((6.0times10^3)timesv^2)/250#
=#(21000times250)/(6.0times10^3)#
3. A stone attached to one end of a string is whirled in space in in a vertical plane. If the length of the string is 80 cm, determine the minimum
speed at which the stone will describe a vertical circle. (Take g =# 10 ms^-2#).
solution:
minimum speed v=#sqrt rg = sqrt0.08 times 10#
=2.283m/s


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