Question

A sphere of radius r=34.5 cm and mass m= 1.80kg starts from rest and rolls without slipping down a 30.0 degree incline that is 10.0 m long. calculate the translational and rotational speed when it reaches the bottom.

Answer 1

from the given data

vertical height, h = 10*sin(30)

= 5 m

Now Apply conservation of energy

final kinetic energy = initial potentail energy

0.5*m*v^2 + 0.5*I*w^2 = m*g*h

0.5*m*v^2 + 0.5*(2/5)*m*r^2*w^2 = m*g*h

0.5*m*v^2 + 0.2*m*r^2*w^2 = m*g*h

0.5*m*v^2 + 0.5*m*v^2 = m*g*h

0.7*m*v^2 = m*g*h

v = sqrt(g*h/0.7)

= sqrt(9.8*5/0.7)

= 8.37 m/s <<<<<<----------Answer

rotational speed, w = v/r

= 8.37/0.345

= 24.3 rad/s <<<<<<----------Answer

angular velocity, w = 4 rev/s

= 4*2*pi rad/s (since 1 revolution = 2*pi radians)

= 25.13 rad/s <<<<<-------Answer