Assuming the average distance from the earth to the sun is 1.48 x 10^11 M and given that the earth revolve around the sun once per year, with what average linear velocity does the earth transverse its orbit? The answer is 29500 m/s but how do you arrive at that answer?
A student attaches a 2.0 kg mass to the end of a 3.5 m long string to form a simple pendulum. What will be the frequency of the pendulum's oscillation? The answer is 0.27 but how do you arrive at that answer?
A thin-walled cylinder with a mass of 23 kg and a radius of 1.4 m is placed atop a 4.5 m tall hill. What will be its linear velocity after it rolls without slipping down the hill? The answer is 6.6 m/s but how do you get to it?
From linear v = circumference/time, v = 2πr/T where T equals one year = 365.25 d * 24 hr * 60 min * 60 s, so v = 2π (1.48 x 1011 m)/(3.156 x 107 s) = 2.95 x 104 m/s.
For a pendulum, 2π f = √(g/L) where g on Earth's surface is about 9.8 m/s2 and L is length in meters. Notice there is no dependence on mass. f = √(9.8/3.5) /(2π) = 0.266 Hz.
Gravitational potential energy at top of hill, mgh, goes into both kinetic energy 1/2 mv2 and rotational energy 1/2 I ω2. Linear and angular velocities are related by v = r*ω; for a hollow cylinder, I = m r2. Then mgh = 1/2 mv2 + 1/2 mv2 because ω2 = v2 / r2. So gh = v2 with mass cancelling out, and v = √(9.8*4.5) = 6.64 m/s.
