Question

Tether ball is a game children play in which a ball hangs from a rope attached to the top of a tall pole. The children hit the ball, causing it to swing around the pole. What is the total initial acceleration of a tether ball on a 2.0 m rope whose angular velocity changes from 13 rad/s to 7.0 rad/s in 15 s

Answer 1

Answer:

a_total = 14.022 m/s²

Explanation:

The total acceleration of a uniform circular motion is given by the following formula:

$a=\sqrt{a_c^2+a_T^2}$         (1)

ac: centripetal acceleration

aT: tangential acceleration

Then, you first calculate the centripetal acceleration by using the following formula:    

$a_c=r\omega^2$

r: radius of the circular trajectory = 2.0m

w: final angular velocity  of the ball = 7.0 rad/s

$a_c=(2.0m)(7.0rad/s)^2=14.0\frac{m}{s^2}$        

Next, you calculate the tangential acceleration. aT is calculate by using:

$a_T=r\alpha$    (2)

α: angular acceleration

The angular acceleration is:

$\alpha=\frac{\omega_o-\omega}{t}$

wo: initial angular velocity = 13 rad/s

t: time = 15 s

Then, you use the expression for the angular acceleration in the equation (1) and solve for aT:

$a_T=r(\frac{\omega_o-\omega}{t})=(2.0m)(\frac{7.0rad/s-13.0rad/s}{15s})=-0.8\frac{m}{s^2}$

Finally, you replace the values of aT and ac in the equation (1), in order to calculate the total acceleration:

$a=\sqrt{(14.0m/s^2)^2+(-0.8m/^2)^2}=14.022\frac{m}{s^2}$

The total acceleration of the ball is 14.022 m/s²