A bob of mass m is suspended from a fixed point with a massless string of length L (i.e., it is a pendulum). You are to investigate the motion in which the string moves in a cone with half-angle θ.

What tangential speed, v, must the bob have so that it moves in a horizontal circle with the string always making an angle from the vertical?
How long does it take the bob to make one full revolution (one complete trip around the circle)?

Express your answer in terms of m, L, θ and acceleration due to gravity g.

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AFOKE88 AFOKE88 · Answer 1 · 2020-03-05T05:04:01+01:00

Answer:

A) Tangential Speed ;

v = √[(tan(θ)) (mgLsin(θ))]

B) period = [2πrLsin(θ)] / √[(tan(θ)) (mgLsin(θ))]

Explanation:

From the question, since the angle from the vertical is θ, then the tension must satisfy the 2 conditions ;

- the vertical component of tension must support the weight of the bob, i.e Tcos(θ)=mg

- the horizontal component of tension must equal the centripetal force acting on the bob, i.e,

Tsin(θ) = (v^(2))/r where r is the radius of the orbit,

However, from geometry, r=Lsin(θ), and so,

Tsin(θ) = (v^(2))/Lsin(θ))

Now, let's divide the horizontal component in the second condition by the vertical component in the first condition above .

So,

[Tsin(θ)]/ [Tcos(θ)] = [(v^(2))/Lsin(θ))]/(mg)

So, We know that sin(θ)/ cos(θ) = tan (θ), thus;

tan(θ) = [(v^(2))/mgLsin(θ))]

So, making v the subject, we have;

v = √[(tan(θ)) (mgLsin(θ))]

Now to find the period time, we know that period = distance / velocity. Now, distance = 2πr

From above, we see that r=Lsin(θ)

Thus, distance = 2πrLsin(θ)

Thus, period = [2πrLsin(θ)] / √[(tan(θ)) (mgLsin(θ))]