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IceJOKER [234]
2 years ago
14

A 4 kg particle moves at a constant speed of 2.5 m/s around acircle of radius 2 m. What is its angular momentum about the center

of the circle? Answerin units of kg · m2/s. What is its moment of inertia about an axis through the center ofthe circle and perpendicular to the plane of the motion? Answer inunits of kg · m2. What is t
Physics
1 answer:
blagie [28]2 years ago
6 0

Explanation:

Given that,

Mass of the particle, m = 4 kg

Speed of the particle, v = 2.5 m/s

The radius of the circle, r = 2 m

We need to find the angular momentum about the center of the circle. The formula for the angular momentum is given by :

L=mvr

Substitute all the values,

L=4\times 2.5\times 2\\\\L=20\ kg{\cdot}m^2s

So, the angular momentum of the particle is 20 kg-m² s.

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Answer:

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A 30-cm-diameter, 1.2 kg solid turntable rotates on a 1.2-cm-diameter, 450 g shaft at a constant 33 rpm. When you hit the stop s
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Answer:

frictional force = 0.52 N

Explanation:

diameter of turn table (D1) = 30 cm = 0.3 m

mass of turn table (M1) = 1.2 kg

diameter of shaft (D2) = 1.2 cm = 0.012 m

mass of shaft (M2) = 450 g = 0.45 kg

time (t) = 15 seconds

acceleration due to gravity (g) = 9.8 m/s^{2}

radius of turn table (R1) = 0.3 / 2 = 0.15 m

radius of shaft (R2) = 0.012 / 2 = 0.006 m

total moment of inertia (I) = moment of inertia of turn table + moment of inertia of shaft

I = 0.5(M1)(R1)^{2} + O.5 (M2)(R2)^{2}

I =  0.5(1.2)(0.15)^{2} + O.5 (0.45)(0.006)^{2}

I = 0.0135 + 0.0000081 = 0.0135081

ω₁ = 33 rpm = 33 x \frac{2π}{60} = 3.5 rad/s

α = -ω₁/t = -3.5 / 15 = -0.23 rad/s^{2}

torque = I x α

torque = 0.0135081 x (-0.23) = - 0.00311 N.m

torque = frictional force x R2

- 0.00311 = frictional force x 0.006

frictional force = 0.52 N

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