Uniform Circular Motion: You need to design a wheel for testing purposes such that its rim will have an acceleration of 1.5 g when the rim is moving at 37 m/s while spinning. What should be the diameter of this wheel?

We want to have an acceleration of:

A = 1.5*g = 1.5*9.8m/s^2  = 14.7 m/s^2

When the rim is moving at 37m/s while spinning.

Now, in circular motion we have two accelerations.

Tangential acceleration, that is the one related to the change of speed, as we have a speed of 37m/s, we can assume that is constant, then the tangential acceleration is zero.

Centripetal acceleration, is the one related to the change in direction, is perpendicular to the velocity vector and is the one that allows the circular motion.

I suppose that in this problem we want to have a  centripetal acceleration of 14.7m/s^2

The equation for the centripetal acceleration is:

Ac = v^2/r

Where v = velocity and r = radius.

then we must solve:

14.7m/s^2 = (37m/s)^2/r

r = (37m/s)^2/14.7m/s^2 = 93.13 m

d

explanation:

v=uf/(u-f)

= 5.10/(5-10) convex lens has positive focal length

=50/-5

=-10 cm (formed at other side)

answer would be letter choice (a),   a ball resting on a shelf. would be another example of potential energy   >   stored energy > not moving.

a ball spinning on a finger   >   kinetic energy > moving energy

a ball rolling on the ground   >   kinetic energy   > moving energy

a   ball flying through the air   > kinetic energy   >   moving energy

hope that                                         : )