After class one day, two physics students get into an argument about rotational

Question

After class one day, two physics students get into an argument about rotational dynamics. Jack thinks the nickel in his pocket would beat jill's silver ring in                     a race down an incline, but Jill is convinced her ring would win. To settle their argument, Jack and Jill climb up a 20 m hill inclined at 30 degrees to the                     horizontal. Both objects are released from rest at the top of the hill and roll without slipping all the way to the bottom. What is the transnational speed of                     each object when it reaches the bottom? How long does each object take to get from the release point to the bottom of the hill?                 

Explanation / Answer

mgh = 0.5*(3/2)*m*v^2

v = sqrt((4/3)*g*h) = sqrt((4/3)*9.8*20)= 16.16 m/s

mgsin30R = (3/2)m*R^2*a/R

gsin30 = (3/2)a

t = v/a = 3V/g = (3*16.16)/9.8 = v s

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