Question

A 6.82 kg bowling ball slipsnand slides down the alley with a center velocity of 7.33 m/s. The radius 10.85 cm ball rotates with an angular velocity of 0.459 rad/s. A) what is the kinetic energy of the ball? B) If the ball rolls along at 7.33 m/s without slipping, what is the kinetic energy?

Answer 1

a) for the ball

kinetic energy = translational kinetic energy + rotational kinetic energy

kinetic energy = 0.50 mv^2 + 0.50 * I * w^2

kinetic energy = 0.50 * 6.82 * 7.33^2 + 0.50 * 2/5 * 6.82 * 0.1085^2 * 0.459^2

kinetic energy = 183.2 J

the kinetic energy is 183.2 J

B)

for rolling

v = r * w

w = v/r

kinetic energy = translational kinetic energy + rotational kinetic energy

kinetic energy = 0.50 mv^2 + 0.50 * I * w^2

kinetic energy = 0.50 mv^2 + 0.50 * 0.40 * m * r^2 * (v/r)^2

kinetic energy = 0.50 mv^2 + 0.50 * 0.40 * m * (v)^2

kinetic energy = 0.70 * m * v^2

kinetic energy = 0.70 * 6.82 * 7.33^2

kinetic energy = 256.5 J