I don\'t understand where 6/5 came from, but the rest is similar to our books ex

Question

I don't understand where 6/5 came from, but the rest is similar to our books example. The teacher's example follows F=mgsin theta - f=ma and then substitutes f=2ma/3 for friction in that equation. The book example makes more sense to me, but I'm not coming up with the correct answer which is time=1.7s and acceleration of 2.94 m/s^2.

A regulation basketball has a 25 cm diameter and may be approximated as a thin spherical shell. The acceleration of gravity is 9:81 m/s2. How long will it take a basketball starting from rest to roll without slipping 4 m down an incline that makes an angle of 30 degrees with the horizontal? Answer in units of s. The given moment of inertia is I = 2/3 MR^2


Energy considerations: ½mv² + ½I?² = mgh
I = (2/3)mr², ? = v/r and h = 4sinT ?
v = v((6/5)*g*4sin30°) = m/s

t = 2x/v = 2*4/velocity = t sec

Explanation / Answer

Never heard of a regulation that has a 25 cm diameter.Let it be R. We have potential energy Pe of the ball gets converted torotaional Ker and traslational kinetic Ket enerrgy. Pe= Ker + Ket Pe=mgh Ker= 0.5I2 Ket= 0.5mV2 Now mgh=0.5I2 + 0.5mV2 since = V/R and I=(2/3)mR2 wehave mgh=0.5 (2/3)mR2 (V/R)2 + 0.5mV2 mgh=0.5 (2/3)m V2 + 0.5mV2 Solving for V we have (2/3) V2 + V2 = 2gh (5/3)V2 = 2gh finally V = [(6/5) gh] finally This is V max or V at athe bottom of the incline V=at also S=0.5at2 ; where S= h/sin(86.0) a= 2S/t2 now since V=at V= [2S/t2 ]t = 2S/t Finally t= 2S/V = 2S/[(6/5) gh] t= 2S [(5/6) /gh] Just plug values, we get the ans.

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