particle in Uniform circular motion question ,,, plz help?

Let's take (c) first because it's easiest. The circumference of a wheel with radius 14m is 2*pi*14 = 88.0 m. If it's moving at 7 m/s. v = d/t ---> t = d/v = 88/7 = 12.6 s

For the first two questions, the acceleration of an object in circular motion is always toward the centre of the circle, so her acceleration will be directly upward at the bottom of the circle (lowest point) and directly downward at the top of the circle (highest point).

The magnitude in both cases will be given by a = v^2/r = 7^2/14 = 3.5 m/s/s

First, what you describe isn't accessible...there are always frictional forces (e.g., drag, kinetic, static, viscosity) appearing on inertial mass of any style. So, there can be something there to decelerate a shifting mass no remember what its structure. 2d, Newt's regulations say once a mass receives shifting, it continues to be shifting interior a similar way (rotation, linear, even if) until eventually a information superhighway rigidity on it differences that action (jointly with bringing it to a halt). So, accepting the no longer accessible (frictionless action), your ring will rotate continually as quickly because this is began. 0.33, centripetal rigidity contained in the ring is strictly offset by technique of centrifugal rigidity using modify in tangential velocity (and hence acceleration) at each "particle" or element contained in the ring. hence, there's no information superhighway rigidity the following to modify the ring's rotational velocity. No, i do not imagine there is something (except truth) to end that ring.

centripetal accleration = v^2/r

a = 7^2 / 14

a = 3.5 m/s^2 towards the centre of the ferris wheel

unless gravity is excluded, value changes depending on place

at lowest point, a = 3.5 - mg

at highest point, a = 3.5 + mg

angular velocity, w

v = w * r

w = v/r

= 7/14

= 0.5 radians per second

2(pi) radians in one revolution

w = (arc length) / (time)

t = (arc length)/ w

t = (2(pi))/ 0.5

t = 4pi seconds