When a small 3.00 g coin is placed at a radius of 5.30 cm on ahorizontal turntab

Question

When a small 3.00 g coin is placed at a radius of 5.30 cm on ahorizontal turntable that makes three full revolutions in 3.34 s,the coin does not slip. What are (a) the coin'sspeed, the (b) magnitude and (c)direction (radially inward-denote 0 or outward-denote 1) of thecoin's acceleration, and the (d) magnitude and(e) direction (inward-denote 0 or outward-denote1) of the frictional force on the coin? The coin is on the verge ofslipping if it is placed at a radius of 10.0 cm.(f) What is the coefficient of static frictionbetween coin and turntable?

Explanation / Answer

(a)       the distance travelled by the coin in 3.14 sis    3 (2 r) = ....... m    so its speed will be    v = ....... m / 3.14 s (b)    the centripetal acceleration is givenby    a = v2 / r       = ....... m /s2 (c)    the acceleration vector is horizontal andpoints from the coin towards the centre of the turn table (d)    the only horizontal force acting on the coinis static friction fs and must be large enough to supplythe acceleration of part    (b) for mass m = 0.030 kg coin    from newtons second law of motion we getthat    fs = m a        = ......... N (e)    the static frictionfs must points in the same direction as theacceleration (f)    from the theory we can see hat the normalforce exerted upwards on the coin by the turntable must equal thecoins weight    so if we repeat parts (a) and (b) for r' =0.14 m    we get a' and v'    if the friction is at its maximum at r = r'then we get that    s = fs,max / mg         = m a' / m g         = ........

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