A student throws a long stick of length 0.11m and mass 0.5 kg intothe air in suc
ID: 1677657 • Letter: A
Question
A student throws a long stick of length 0.11m and mass 0.5 kg intothe air in such a way that the center of the mass rises vertically.At the moment it leaves his hand the stick is horizontal and thespeed of the end of the stick is nearest to him is zero. When thecenter of mass of the stick reaches its highest point, the stick ishorizontal, and it has made 26 complete revolutions.How long did it take for the center of mass to reach its highestpoint? Assume the stick's cross sectional area and mass is uniform.The acceleration of gravity is 9.8m/s^2.Explanation / Answer
1. The rotational Velocity of the stick () is constant butthe vertical velocity is effected by gravity. 2. Initial velocity of center of mass v.i = * r r = 1/2 length Note: This means the velocity at the hand = 0 and the velocity atthe far end of the stick is = 2*v.i 3. The time it takes to do 15 rotations is t = 15 * 2 / 4. But we can also know that t= v.i/g t = r /g Combine equations 3 and 4 t = r /g = 15 * 2 / ^2 = g/r * 15 *2 therefore = (30 g/r) Accordingly t = r/g t = (30 g/r) * r/g t = (30 r/g) r= L/2 t = (15 L/g) t = 0.726 secs ********** Captain M is one of my buddies on Y!A and I always presume that heis correct unless shown otherwise. (You will note that both theCaptain and I use the same basic methodology). Here, Captain M erred on his formula for T. Let me quote him: "We are told that the stick completes 15 rotations. Use the speedof the end of the stick to find the time of one rotation,remembering that the stick is rotating around the center of thestick (giving a radius of the length of the stick divided by2). Time of one rotation = T = circumference/(speed of the end)" *No. It is not the absolute speed of the end, it is the speed ofthe end relative to the speed of the center. This is because 1/2 ofthe speed of the end is in rotation and the other 1/2 is invertical velocity. (I hope I made this clear, it can bedifficult). *This makes his next formula T = 2*pi*r/v = 2*pi*(L/2)/(2*V0 - V0) = pi*L/(V0) *Continuing with the correct formula and using Captain M'sanalysis: Time to max height = t = 15*T = 15*pi*L/(V0) 0 = V0 - g[15*pi*L/(V0)] V0 = SQRT[g*15*pi*L] ... g = 9.8 b/s^2 , L = 0.11 m V0 = 7.13 m/s We can now use this to get the time to max height. t = 15*pi*L/(V0) t = 0.727 seconds
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