## Point P is located on the outer surface of the top, a distance h = 29.0 mm h=29.0 mm above the ground. The angle that the outer surface of t

Question

Point P is located on the outer surface of the top, a distance h = 29.0 mm h=29.0 mm above the ground. The angle that the outer surface of the top makes with the rotation axis of the top is θ = 19.0 ∘ . θ=19.0∘. If the final tangential speed v t vt of point P is 1.45 m/s, 1.45 m/s, what is the top’s moment of inertia I ?

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Physics
7 mins
2021-07-22T20:27:14+00:00
2021-07-22T20:27:14+00:00 1 Answers
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## Answers ( )

Answer:

Explanation:

Given that,

Height h = 29mm = 0.029m

Makes an angle of θ = 19.0°

Then, the radius is

h = rCosθ

r = h/Cosθ

r = 0.029/Cos19

r = 0.031m

Linear speed v = 1.45m/s

The relationship between linear speed and angular speed is

v = ωr

ω = v/r

ω = 1.45/0.031

ω = 47.28 rad/sec

Angular momentum is given as

L = Iω

Therefore, I = L / ω

So, the question is not complete check attachment for the first part of the question, from the first part we will get the angular momentum

Given that,

r = 3.46mm = 0.00346m

Tension in string F = 3.15N

Time taken t = 0.62s.

Then, Torque is given as

T = dL/dt

Tdt = dL

Integrate both sides

T•t + C = L(t)

At t = 0 L=0

Therefore C = 0

T•t = L(t)

The torque from the two string is given as

T = 2Fr

Therefore,

L(t) = 2•F•r•t

L(t) = 2 × 3.15 × 0.00346 × 0.62

L(0.62) = 0.0135 kgm²/s

Then, back to our question,

I = L / ω

I = 0.0135/47.28

I = 2.86 × 10^-4 kgm²