(a) 6.04 rev/s
The speed of the ball is given by:
where
is the angular speed
r is the distance of the ball from the centre of the circle
In situation 1), we have
r = 0.600 m
So the speed of the ball is
In situation 2), we have
r = 0.900 m
So the speed of the ball is
So, the ball has greater speed when rotating at 6.04 rev/s.
(b)
The centripetal acceleration of the ball is given by
where
v is the speed
r is the distance of the ball from the centre of the trajectory
For situation 1),
v = 30.6 m/s
r = 0.600 m
So the centripetal acceleration is
(c)
For situation 2 we have
v = 34.1 m/s
r = 0.900 m
So the centripetal acceleration is
To solve this problem we will apply the concepts related to the centripetal force, for which it is necessary to equate it with the static friction force of the body. From this, we will clear the speed and replace with the given values. Our values are defined as,
Maximum velocity can be find out using centripetal force,
Must be equal to,
Therefore the maximum speed that he can travel through the arc without slipping is 9.93m/s
Answer:
When a custodian pushes a large crate across the gym floor, forces acting on the crate are the one applied by the custodian, frictional force acting in the opposite direction of motion acting between the surface of crate and floor and reaction force (normal force) acting in upward direction normal to the surface.
where is the gravitational constant and is about
:D
Work is calculated by multiplying the force exerted with the distance covered by the object. The force and the displacement should be parallel. In this item, we have to calculate first the horizontal component of force.
x-component of force = (26 lbs)(cos 30°)
= 22.52 lbs
Multiplying the calculated value by the displacement of 90 ft will give us the final answer of 2026.50 lb.ft.