1.) The mass of the sun is 329 320 times that of the earth. The earth-sun separa

Question

1.) The mass of the sun is 329 320 times that of the earth. The earth-sun separation measured from their centers is about 1.496 x10^11 m. Determine the location of the center of mass of the earth-sun systemas measured from the center of the sun.
a.) 454,171 m b.) 416,957 km c.) 4,928,621 km d.) 1,061 m e.) 86,000 m

2.) The fact that the Earth has lost about 35 days per year over the past 400 million years is a consequence of the Earth's:
a.) Centripetal deceleration b.) Rotational angular deceleration c.) Changing Moment of Inertia d.) Increasing rotational angular velocity e.) Increased orbital velocity about the sun

3.) A wheel of radius 12 cm is mounted on a frictionless, horizontal axle that is perpendicular to the wheel and passes through the wheel's center of mass. A light cord wrapped around the wheel supports a 0.40 kg weight. When released from rest with the string taut, the weight accelerates downward at 3.0 m/s^2. What is the wheel's moment of inertia?
a.) 0.035 kg m^2 b.) 0.013 kg m^2 c.) 0.023 kg m^2 d.) 0.020 kg m^2 e.) 0.016 kg m^2

Explanation / Answer

3)

In the equation, net torque = I *a, 'a' means angular acceleration (rad/s^2) not the linear acceleration of the falling object (3.0m/s^2). You have also calculated the torque incorrectly - the force is the tension, not the weight.

Resultant force on falling mass F = ma = 0.40 x 3.0 = 1.2N
Weight of mass, W = mg = 0.40 x 9.8 = 3.92N
If tension in rope is T, then F = W - T, so T = W - F = 3.92 - 1.2 = 2.72N
Torque on wheel = Txd = 2.72x0.12 = 0.3264Nm

For angular velocity w, the linear velocity, v, is given by v = wr. It follow (from calculus if you know it dv/dt = r*dw/dt) that

linear acceleration (at edge of wheel) = r x angular acceleration

Angular acceleration = linear acceleration / r = 3.0/0.12 = 25rad/s^2

I = torque/ a_angular = 0.3246/25

= 0.013kgm^2

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