Question

QUESTION 1 A bowling ball rolls 2.3 m up a ramp without slipping. It has an initial speed of its center of mass of 7.9 m/s, and the ramp is 20.3 degrees up from the horizontal. What is its speed at the top of the ramp? (HIN T: Consider which approach is the best to use here: Kinematics+Dynamics or Energy?)

Answer 1

given

v1 = 7.9 m/s
L = 2.3 m
theta = 20.3 degrees
let m is the mass and r radius of the ball.
v2 = ?
vertical height reached, h = L*sin(theta)

= 2.3*sin(20.3)

= 0.798 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

(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(7.9^2 - 10*9.8*0.798/7)

= 7.16 m/s <<<<<<<<<<<<-----------------------Answer

Please comment for further help on this problem.