Question

A spherical shell is released from rest and rolls down a 2 = 28° incline without slipping and reaches the bottom with an angular speed of w = 32.2 rad/s. The M = 1.5 kg sphere has a radius R = 0.60 m and a moment of inertia given as I = (2/3)MR2. R -AX 0 Find the distance Ax that the sphere traveled on the incline. m

Answer 1

Mass of the sphere, M=1.5 kg; Radius, R=0.60m Morneut y the inertia of lte spherical shell, I = Me’s 2x15x0.62 I=0.36 kg m² Angular speed at the base of the incline, W = 32.2 radis Angle of incline, o=28° Initial potential energy of the shell, Ep = mgh where he vertical height Total Kine tr energy of the shell at the base of the incline, Ex= Etrans + Egit =IMV²+ L I w² ; But V=Rw IMR?W? + 1x2 MR2 w² MR²W 5 MR²w2 Using law of conservation of energy Ep = Ex Mgh=5 MR² W² => b=5R²W - 5x0.60²x 32.22 69 2 2 2. Ex= (1+2) = 8 2. 2 3 - 3174 m 6 6x9-8 If ay is the distance travelled by the sphere along the incline, Sino=h > BX = hylsing 31.74f51b28 = 676m distance travelled by the sphere along the incline, AX = 67.6 m АХ