A 300 ball swings in a vertical circle at the end of a 1.2-m-long string. When the ball is at the bottom of the circle, the tension in the string is 13 .
What is the speed of the ball at that point?
tension at bottom= mg+mv^2/r solve for v.
Wass goodie add da snap @brownieboe
Someone answer the question with a detailed description and contact me via snap ^^^
I mean its due the 12th and no one will probably see this for like 5 years... if your from Mrs Lovelands class tell her I said even if she dont remeber me in the future.
^doesnt
To find the speed of the ball at the bottom of the circle, we can use the concept of centripetal acceleration. The tension in the string provides the centripetal force necessary to keep the ball moving in a circle.
The centripetal force (F) is given by the equation:
F = m * a
Where:
m = mass of the ball = 300 grams (or 0.3 kg)
a = centripetal acceleration
The centripetal acceleration (a) is given by the equation:
a = v^2 / r
Where:
v = velocity of the ball
r = radius of the circle = 1.2 m
The tension in the string (T) is equal to the centripetal force (F). So, we have:
T = F
Plugging in the values, we get:
13 = m * a
13 = 0.3 * a
Dividing both sides by 0.3, we get:
a = 43.33 m/s^2
Now, we can substitute this value of centripetal acceleration into the centripetal acceleration equation:
a = v^2 / r
43.33 = v^2 / 1.2
Multiply both sides by 1.2:
51.996 = v^2
Taking the square root of both sides:
v = √51.996
v ≈ 7.21 m/s
Therefore, the speed of the ball at the bottom of the circle is approximately 7.21 m/s.