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Elina [12.6K]
2 years ago
11

A boy is swinging a toy on a piece of string in a vertical circle. The toy has a mass of 150 g and the radius of the circle is 0

.8 m. a) He swings the toy with a linear velocity of 2 m/s. Will the toy move in a circle? Explain your answer. b) Another boy swings the toy with a linear velocity of 3.5 m/s. Work out the tension in the string at the top of the circle, at the bottom of the circle and halfway between the top and the bottom of the circle.​
Physics
1 answer:
Maru [420]2 years ago
3 0
At the top of the circular motion, both weight and tension provides for centripetal force.

By Newton’s Second Law,
Fnet = ma
mg + T = mv^2/r (since a = v^2/r and weight = mg)

For toy to continue moving in circle at the top,

T > 0
mv^2/r - mg > 0
v >root (gr)

Hence, minimum speed toy must have is 2.80 m/s. Since linear velocity is lower than the minimum linear velocity, the toy will not move in circular motion.

b) Tension at top = mv^2/r - mg
= (0.15)(3.5)^2/0.8 - (0.15)(9.81)
= 0.825 N

Tension at bottom = mv^2/r + mg
= (0.15)(3.5)^2/0.8 + (0.15)(9.81)
= 3.77 N

In the middle, only Tension provides for centripetal force. Hence,
Tension = mv^2/r
= (0.15)(3.5)^2/0.8
= 2.30 N
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Falling raindrops frequently develop electric charges. Does this create noticeable forces between the droplets? Suppose two 1.8
umka2103 [35]

Answer:

a) F=2.048\times 10^{-7}\ N

b) a=0.1138\ m.s^{-2}

Explanation:

Given:

  • mass of raindrops, m=1.8\times 10^{-6}\ kg
  • charge on the raindrops, q=+21\times 10^{-12}\ C
  • horizontal distance between the raindrops, r=0.0044\ m

A)

<u>From the Coulomb's Law the force between the charges is given as:</u>

F=\frac{1}{4\pi.\epsilon_0} .\frac{q_1.q_2}{r^2}

we have:

\epsilon_0=8.854\times 10^{-12}\ C^2.N^{-1}.m^{-2}

<em>Now force:</em>

F=\frac{1}{4\pi\times 8.854\times 10^{-12}} .\frac{21\times 10^{-12}\times 21\times 10^{-12}}{0.0044^2}

F=2.048\times 10^{-7}\ N

B)

<u>Now the acceleration on the raindrops due to the electrostatic force:</u>

a=\frac{F}{m}

a=\frac{2.048\times 10^{-7}}{1.8\times 10^{-6}}

a=0.1138\ m.s^{-2}

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