Physics: Vertical Centripetal Acceleration?

Question

A 900 g ball moves in a vertical circle on a 1.09 m-long string. If the speed at the top is 3.90 m/s, then the speed at the bottom will be 7.61 m/s.

What is the ball's weight?

What is the tension in the string when the ball is at the top?

What is the tension in the string when the ball is at the bottom?

Answer 1

All you need is two formulas.
Part i
Weight=mass x (acceleration due to gravity)
or W=mg
m is given to you in the question- .9(it must be in SI units i.e. Kg's)
g=9.8(standard figure to one decimal place on earths surface)
so W=(.9)(9.8)
W=8.82N

Part ii
for this part you need to use the formula
Centriptial Force(i.e. in this case Tension)=mass*velocity(squared)/radius of motion
T=mv^2/r
T=(.9)(3.9)^2/1.09
T=13.689/1.09
T=12.5587
Since values are given to 2 decimal places in question you can round off to two decimal places in answer
T=12.56

Part iii
Same formula again
T=(.9)(7.61)^2/1.09
T=47.82

Answer 2

> What is the ball's weight?

That one's easy. It's (900g) × (g).

> What is the tension in the string when the ball is at the top?

Use these equations:
Fnet = tension + weight (both downward)
Fnet = ma
a = v²/r

They give you "v", "r" and "m"; and you've already calculated "weight". Combine the above 3 equations, and solve for "tension".

> What is the tension in the string when the ball is at the bottom?

Just like the previous question, except this time the weight and the tension act in opposite directions, so you subtract them instead of adding them:

Fnet = tension − weight (tension "up"; weight "down")

As before, combine that with these:
Fnet = ma
a = v²/r

Remember that you have a different value of "v" this time. Combine equations and solve for "tension."

Answer 3

There are 2 forces exerting on the ball: the stress stress of the rope and the gravity. the stress stress (radial) is often favourite to the cost (tangential) of the ball, consequently the paintings by utilising the stress equals 0. The paintings achieved by utilising the gravity whilst the ball is going from the extraordinary to the backside of the circle is W = 2MgR That paintings will advance the kinetic ability of the ball from KE(extraordinary) = M v1²/2 to KE(backside) = Mv2²/2 KE(backside) - KE(extraordinary) = W M v2² / 2 - M v1² / 2 = 2 M g R v2² - v1² = 4g R (V2² / R) - (V1² / R) = 4 g a2 - a1 = 4 g a2 = a1 + 4 g = 3 g + 4 g = 7 g

Answer 4

Shoutout to Dr. Wang

Answer 5

weight is the mass times g

tension at the top

http://img458.imageshack.us/img458/3434/tensiontopofcxs5.jpg

tension at the bottom

http://img458.imageshack.us/img458/7078/tensionbottomofcao9.jpg