Question

A solid ball is released from rest at the top of a 1.60 m -long ramp inclined at 18 degrees. At the bottom, the ball continues along a flat section that's also 1.60 m long.

Whats the overall travel time?

Answer 1

Using energy conservation

KEi + PEi = KEf + PEf

KEi = 0, since velocity is zero

PEf = 0 since h = 0

So,

mgh = 0.5*Iw^2 + 0.5*mv^2

I = 0.4*m*r^2

w = v/r

using above values

mgh = 0.5*0.4m*r^2*(v/r)^2 + 0.5*m*v^2

mgh = 0.7*m*v^2

v = sqrt (2*g*h)

h = 1.6*sin 18 deg = 0.494 m

v = sqrt (2*9.81*0.494) = 3.113 m/sec

Vf = Vi + a*t1

a = g*sin 18 deg

t1 = Vf/a = 3.113/(9.81*sin 18 deg) = 1.026 sec

B.

V = 3.113 m/sec

d = 1.6

t = 1.6/3.113 = 0.513 sec

Total time = t1 + t = 1.026 + 0.513 = 1.539 sec