Question

A solid ball of diameter 20.0 cm is rolling without slipping to the right as shown below. The initial period of the ball is 1.00 s. The height of the ramp is h = 0.500 m and it is at an angle of 40.0°.

A) What is the period of the ball at the bottom of the incline?

The ball now rolls without slipping up a ramp on the right that has an incline of 20.1°

B) How high up the ramp on the right will the ball go?

40.0 20.1°

Answer 1

A)
d = 20 cm = 0.2 m
r = d/2 = 0.2/2 = 0.1 m
T1 = 1 s

w1 = 2*pi/T1 = 2*pi/1 = 6.28 rad/s

let m is the mass of ball.

moment of inertia of the ball, I = (2/5)*m*r^2

v1 = r*w1 = 0.1*6.28 = 0.628 m/s

let v2 is the linear and w2 is the angular speed at the bottom.

Apply conservation of energy

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

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

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

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

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

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

= sqrt(0.628^2 + 10*9.8*0.5/7)

= 2.72 m/s

w2 = v2/r

= 2.72/0.1

= 27.2 rad/s

T2 = 2*pi/w2

= 2*pi/27.2

= 0.231 s <<<<<<<<<<<<<------------------------Answer

B) let H is the vertical height reached on second inclined path.

Apply conservation of energy

initial mechanical energy = final potential energy

(7/10)*m*v2^2 = m*g*H

==> H = 7*v2^2/(10*g)

= 7*2.72^2/(10*9.8)

= 0.528 m

distance travelled along the ramp, L = H/sin(20.1)

= 0.528/sin(20.1)

= 1.54 m <<<<<<<<<<<<<<-----------------Answer

if we need only vertical height, 0.528 will be the answer.