A 0.45-kg ball, attached to the end of a horizontal cord, is rotated in a circle of radius 1.3-m on a frictionless horizontal surface.
If the maximum speed of the ball is 14.7 m/s, how long does it take the ball to make on rotation if it is moving at its maximum speed?
circumference = 2 pi R
time in seconds
= distance / speed = 2 pi (1.3) /14.7
To solve this problem, we can use the formula for the centripetal force:
F = m * (v^2 / r)
where F is the centripetal force, m is the mass of the ball, v is the velocity of the ball, and r is the radius of the circular path.
In this case, the maximum speed of the ball is given as 14.7 m/s and the radius of the circular path is given as 1.3 m. We need to find the time it takes for the ball to make one rotation at this maximum speed.
The centripetal force is provided by the tension in the cord that is attached to the ball. This tension is equal to the centripetal force, so we have:
T = m * (v^2 / r)
We also know that the tension is related to the speed of the ball by the formula:
T = mv^2 / r
We can rearrange this equation to solve for time (t):
t = 2πr / v
Plugging in the values we have:
t = 2π * 1.3 m / 14.7 m/s
t ≈ 6.672 seconds
Therefore, it takes approximately 6.672 seconds for the ball to make one rotation at its maximum speed.
Ohhhhhhh...
so we use the CIRCUMFERENCE, not the RADIUS. Got it. Thank you so much!