A 160g stone, tied to a thread, swings in a horizontal circle with a 100 cm diameter. If the maximum tension the thread can withstand before snapping is 28n, find the A) Maximum speed of the stone. B) Acceleration of the stone at this speed. For A I used V=sqrt (r)(F)/m and come up with 9.35 m/s. On part B if I use a=v^2/r I come up with 174.84 m/s^2. This is a crazy speed. Is B a trick question where there is no acceleration. The only other formulas I know are a=delta t/delta v. I would appreciate any assistance. Thanks.

Question

A 160g stone, tied to a thread, swings in a horizontal circle with a 100 cm diameter. If the maximum tension the thread can withstand before snapping is 28n, find the A) Maximum speed of the stone. B) Acceleration of the stone at this speed. For A I used V=sqrt (r)(F)/m and come up with 9.35 m/s. On part B if I use a=v^2/r I come up with 174.84 m/s^2. This is a crazy speed. Is B a trick question where there is no acceleration. The only other formulas I know are a=delta t/delta v. I would appreciate any assistance. Thanks.

Answer 1

tensionmax=mv^2/r

You are correct in both counts, the way to work it, and your answers. Yes, it is crazy.

Answer 2

God Bless you! I have been agonizing over this for some time. Thank you so much!

Answer 3

Let's break down the problem step by step.

A) Maximum Speed of the Stone:
To find the maximum speed of the stone, we can use the formula `v = sqrt((r * F) / m)`, where:
- v represents the speed,
- r represents the radius (half of the diameter of the circle),
- F represents the tension in the thread, and
- m represents the mass of the stone.

Given:
- Diameter = 100 cm, which gives us a radius of 50 cm (or 0.5 m).
- Tension in the thread (F) = 28 N.
- Mass of the stone (m) = 160 g = 0.16 kg.

Now we can substitute these values into the formula and calculate the maximum speed:
v = sqrt((0.5 * 28) / 0.16)
v = sqrt(14 / 0.16)
v = sqrt(87.5)
v ≈ 9.35 m/s

So, you were correct in your calculation. The maximum speed of the stone is approximately 9.35 m/s.

B) Acceleration of the Stone at Maximum Speed:
To find the acceleration of the stone, we can use the centripetal acceleration formula `a = v^2 / r`, where:
- a represents the centripetal acceleration.

Given:
- Maximum speed (v) = 9.35 m/s
- Radius (r) = 0.5 m

Now we can substitute these values into the formula and calculate the acceleration:
a = (9.35)^2 / 0.5
a = 87.5225 / 0.5
a = 175.045 m/s^2

So, the acceleration of the stone at maximum speed is indeed approximately 175.045 m/s^2. It seems like a very high value because it represents the acceleration towards the center of the circle.

Generally, it is not a trick question. In this case, the acceleration is high due to the stone's speed and the small radius of the circular path. The acceleration is necessary to keep the stone moving in a circular path.