A solid sphere is released from the top of a ramp that is at a height It goes do

Question

A solid sphere is released from the top of a ramp that is at a height

It goes down the ramp, the bottom of which is at a height of

above the floor. The edge of the ramp is a short horizontal section from which the ball leaves to land on the floor. The diameter of the ball is 0.14 m.

A solid sphere is released from the top of a ramp that is at a height h1 = 2.05 m. It goes down the ramp, the bottom of which is at a height of h2 = 1.62 m above the floor. The edge of the ramp is a short horizontal section from which the ball leaves to land on the floor. The diameter of the ball is 0.14 m. Through what horizontal distance d does the ball travel before landing? Consider the motion along the ramp first. What is the speed of the ball at the bottom of the ramp? Now consider the motion of the ball after it leaves the ramp. What is the path of the ball once it leaves the ramp? m How many revolutions does the ball make during its fall? What is the angular speed of the ball at the moment it leaves the ramp? How is angular speed defined in terms of the time and the angle? rev

Explanation / Answer

let v is the speed of ball at height h2

according to consrvation of energy

m*g*(h1-h2) = 0.7*m*v^2

v = sqrt(g*(h1-h2)/0.7)

= sqrt(9.8*(2.05-1.62)/9.8)

= 0.656 m/s

angular speed of the when it falls

w = v/r = 0.656/0.07 = 9.368 rad/s

let t is the time taken to fall down from h2 height

h2 = 0.5*g*t^2

t = sqrt(2*h2/g)

= sqrt(2*1.62/9.8)

= 0.574 s

a) d = v*t = 0.656*0.574 = 0.377 m

b)

theta = w*t = 9.368*0.574 = 5.377 radians

= 0.85624 revolutions

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