Question
A coin of mass m = 2.0 g is placed at a distance r = 10.0 cm from the center of a turntable. The coefficient
A coin of mass m = 2.0 g is placed at a distance r = 10.0 cm from the center of a turntable. The coefficient of static friction between the coin and the turntable is ?s = 0.45. Starting from rest, the turntable is very gradually rotated faster and faster. At what angular velocity does the coin slip and "fly off"?
a) Begin with a pictorial representation. Draw the turntable both as seen from above and as an edge view with the coin on the left side coming toward you. Label the radius r, make a table of known information, and indicate what you are trying to find.
b) What direction does the force of static friction point? Explain.
c) What condition describes the situation just as the coin starts to slip? Write this condition as a mathematical statement.
d) Now draw a free-body diagram of the coin. Draw the free-body diagram with the circle viewed edge-on, the r-axis pointing toward the center of the circle, and the z-axis perpendicular to the plane of the circle. Your free-body diagram should have three forces.
e) Write Newton's 2nd Law for the r- and z-directions of the forces. One sum should equal zero and the other should equal m v2/r.
f) The two equations of part e are valid for any angular velocity up to the point of slipping. If you combine these with your statement from part c, you can solve for the speed, vmax, at which the coin slips. Do so.
Newton 2nd law Fnet = ma here 9 = centribetul acceleration = 2 2 8 N= mg ( No motion alon z- direction ( fs ) = mu ? ( along radially . - (2) us mg = mu 2 (f ) Egh Go N = mg fs = MSN from equation HSN = muz Hs mg = m vmax 2 Vmax = Jus79 Vmax = 1 0145 x 0 . 1x9.8 Vmax = 0 66 41 mls this is maximum velocity for which the coil will be in sest for 8 = 0.I mStep by Step Solution
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