Help finding mass and center of mass?

A student throws a long stick into the air in such a way that the center of mass rises vertically.?

• A student throws a long stick of length 0.11m and mass 0.5 kg into the air in such a way that the center of mass rises vertically. At the moment it leaves his hand, the stick is horizontal and the speed of the end of the stick nearest to him is zero. When the center of mass of the stick reaches its highest point, the stick is horizontal, and it has made 15 complete revolutions. How long did it take for the center of mass to reach its highest point? Assume the stick's cross sectional area and mass is uniform. The acceleration of gravity is 9.8 m/s^2. Answer in units of s. I am so confused on this one. I don't understand which equations fit this problem or even how to go about finding them... Help?

• Answer:

  This is an interesting question. It is difficult because you need to marry a bunch of divergent concepts. 1. The rotational Velocity of the stick (ω) is constant but the vertical velocity is effected by gravity. 2. Initial velocity of center of mass v.i = ω * r r = 1/2 length Note: This means the velocity at the hand = 0 and the velocity at the far end of the stick is = 2*v.i 3. The time it takes to do 15 rotations is t = 15 * 2π / ω 4. But we can also know that t= v.i/g t = ω r /g Combine equations 3 and 4 t = ω r /g = 15 * 2π / ω ω^2 = g/r * 15 *2 π therefore ω = √ (30π g/r) Accordingly t = ω r/g t = √ (30π g/r) * r/g t = √ (30π r/g) r= L/2 t = √ (15π L/g) t = 0.726 secs ********** Captain M is one of my buddies on Y!A and I always presume that he is correct unless shown otherwise. (You will note that both the Captain 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 speed of the end of the stick to find the time of one rotation, remembering that the stick is rotating around the center of the stick (giving a radius of the length of the stick divided by 2). Time of one rotation = T = circumference/(speed of the end)" *No. It is not the absolute speed of the end, it is the speed of the end relative to the speed of the center. This is because 1/2 of the speed of the end is in rotation and the other 1/2 is in vertical velocity. (I hope I made this clear, it can be difficult). *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's analysis: 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 *With the correction, Captain M and I come up with the same answer. *********** The difficulties with this question are: a. The velocity of the rotating stick. At the moment of release, the velocity caused by the rotation down of the end of the stick in the thrower's had is exactly equal and opposite to the overall (non-rotational) velocity of the stick going up. This makes the net velocity in the hand = 0 b. The result is arbitrary. There is no way to easily imagine it. You just have to realize that there is a relationship between initial velocity (v.i), rotational velocity (ω), gravity (g), and time (t) and/or height (h) until the stick reaches it's heighest point. You then have to plug and chug until you solve for the unknown. It is like doing a sudoku puzzle. Most problems are a little easier to imagine what the answer should be