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? How many revolutions does the ball make during its fall?

Explanation / Answer

solid sphere diameter d = 0.14 m

radius r = 0.07 m

height h1 = 2.05 m

height h2 = 1.62 m

a)

by using conservation of energy,

P.E= K.E

m*g*h1=1/2mv^2+1/2Iw^2

but,

I = 2/5 mr^2

=1/2mv^2+1/2(2/5)mr^2*w^2

= 1/2mv^2+1/5m*v^2 ( r = v / ? )

= 7/10*m*v^2

so,

m*g*h1 = 7/10*m*v^2

g*h1 = 7/10*v^2

==>

v=sqrt(10/7*g*h1)

=sqrt(10/7*9.8*2.05)

=5.357 m/sec

now.

h2 = 1/2*g*t^2

t=sqrt(2*h2/g)

=sqrt(2*1.62/9.8)

=0.5749 sec

now,

distance d=v*t

=5.357*0.5749

=3.0802155 m ...is answer

b)

theta=w*t

=(v/r)*t

=vt/r

=d/r

=3.0802155 m/0.07 m

=44.0030797 radians

or

=44.0030797/2*pi rev

=7.0068 rev ...is answer

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