A child swings a tennis ball attached to a 0.662-m string in a horizontal circle above his head at a rate of 4.50 rev/s.
(a) What is the centripetal acceleration of the tennis ball?
m/s2
omega^2 R
omega = 4.5 revs/s * 2 pi radians/rev
= 9 pi rad/s
omega^2 = 81 pi^2 /s^2
so
a = 81 pi^2 * .662
= 529 m/s^2
To find the centripetal acceleration of the tennis ball, we can use the formula:
ac = (v^2) / r
Where:
ac = centripetal acceleration
v = tangential velocity
r = radius of the circular motion
First, let's calculate the tangential velocity:
v = 2πr × f
Where:
v = tangential velocity
r = radius of the circular motion
f = frequency in rev/s
Given that the frequency is 4.50 rev/s and the radius is 0.662 m, we can calculate the tangential velocity:
v = 2π(0.662 m) × 4.50 rev/s
Simplifying this equation, we get:
v = 14.19 m/s
Now, substituting the value of the tangential velocity and the radius into the formula for centripetal acceleration:
ac = (14.19 m/s)^2 / 0.662 m
Simplifying this equation, we get:
ac = 321.33 m/s^2
Therefore, the centripetal acceleration of the tennis ball is 321.33 m/s^2.
To find the centripetal acceleration of the tennis ball, we can use the formula:
\[a_c = \frac{v^2}{r}\]
where \(a_c\) is the centripetal acceleration, \(v\) is the velocity of the tennis ball, and \(r\) is the radius of the circular path.
First, we need to find the velocity of the tennis ball. The velocity can be calculated using the formula:
\[v = 2\pi f r\]
where \(f\) is the frequency (in revolutions per second) and \(r\) is the radius of the circular path.
Given that the frequency \(f\) is 4.50 rev/s and the radius \(r\) is 0.662 m, we can substitute these values into the formula to find the velocity:
\[v = 2\pi(4.50 \, \text{rev/s})(0.662 \, \text{m})\]
Performing the calculation gives:
\[v \approx 8.805 \, \text{m/s}\]
Now that we have the velocity \(v\) and the radius \(r\), we can substitute these values into the formula for centripetal acceleration:
\[a_c = \frac{(8.805 \, \text{m/s})^2}{0.662 \, \text{m}}\]
Calculating this gives:
\[a_c \approx 117.53 \, \text{m/s}^2\]
Therefore, the centripetal acceleration of the tennis ball is approximately 117.53 m/s².