Question

A bowling ball rolls 1.9 m up a ramp without slipping. It has an initial speed of its center of mass of 5.3 m/s. and the ramp is 20.8 degrees up from the the horizontal. What is its speed at the top of the ramp?

Answer 1

given

v1 = 5.3 m/s
L = 1.9 m
theta = 20.8 degrees
let m is the mass and r radius of the ball.
v2 = ?
vertical height reached, h = L*sin(theta)

= 1.9*sin(20.8)

= 0.6747 m

let w1 is the initial angular speed and w2 is the final angular speed.

Apply conservation of energy

final kinetic energy + potential energy = initialkinetic energy

(1/2)*m*v2^2 + (1/2)*I*w2^2 + m*g*h = (1/2)*m*v1^2 + (1/2)*I*w1^2

(1/2)*m*v2^2 + (1/2)*(2/5)*m*r^2*w2^2 + m*g*h = (1/2)*m*v1^2 + (1/2)*(2/5)*m*r^2*w1^2 (since v = r*w)

(1/2)*m*v2^2 + (1/2)*(2/5)*m*v2^2 + m*g*h= (1/2)*m*v1^2 + (1/2)*(2/5)*m*v1^2

(7/10)*m*v2^2 + m*g*h = (7/10)*m*v1^2

v2^2 + 10*g*h/7 = v1^2

v2 = sqrt(v1^2 - 10*g*h/7)

= sqrt(5.3^2 - 10*9.8*0.6747/7)

= 4.32 m/s <<<<<<<<<<<<-----------------------Answer