Question

1) Domingo drove his motorcycle at the velocity of 30m/s in 10 meter radius hemispherical hill. What is the centripetal acceleration of Domingo as he passes the top of the hill?

2) An Object slid from a ramp of θ=45ᵒ, starting from rest. The length of the ramp is 100 meter and has no friction.

a)What is the acceleration of the object sliding on the ramp?

b) What is the velocity of the ramp at the bottom of 100 meters?

c)How did the object took to reach to the bottom of the ramp?

Answer 1

Given

1)

velocity v = 30 m/s , radius r = 10 m

the centripetal force is f = mv^2/r

ma =mv^2/r

a = v^2/r  

a = 30^2/10 = 90 m/s2

centripetal acceleration is a = 90 m/s2

2)

theta = 45 degrees

length of the ramp is s = 100 m

a)

acceleration of he object sliding the ramp is  

ma = g sin theta

a = g sin theta

a = 9.8 sin45 m/s2

a= 6.93 m/s2

b) velocity of the ramp at the bottom is  

conservation of energy mgh = 0.5*m*v^2

v = sqrt(2*g*h)

v= sqrt(2*g*s sin theta)

v = sqrt(2*9.8*100 sin45) m/s

v = 37.23 m/s

c)

from equations of motions  

s = ut +0.5*at^2

s = 0 + 0.5*a*t^2

100 = 0.5*6.93*t^2

solving for t , t = 5.37 s