Answer to Question #92895 in Mechanics | Relativity for Ugochukwu

Question

You are playing tetherball with a friend and hit the ball so that it begins to travel in a circular horizontal path. If the ball is 1.2 meters from the pole, has a speed of 3.7 m/s, a mass of 0.42 kilograms, and its (weightless) rope makes a 49° angle with the pole, find the tension force that the rope exerts on the ball just after you hit it.

Answer 1

Consider the schematic figure:

The 2nd Newton's law applied to the current problem states:

$m\vec{a} = m \vec{g} + \vec{T}$

According to the conditions given, the centripetal force is horizontal. Thus, making the horizontal projection of the equation above, we obtain:

$m a_{c.p.} = T \sin{\alpha} \quad \Rightarrow \quad T = \frac{m a_{c.p.}}{\sin{\alpha}}$

The centripetal acceleration can be calculated as:

$a_{c.p.} = \frac{v^2}{r} = \frac{v^2}{l \sin{\alpha}}$

Finally, we derive:

$T = \frac{m v^2}{l \sin^2{\alpha}}$

Substituting the numerical values, we obtain:

$T = \frac{0.42 \cdot 3.7^2}{1.2 \cdot \sin^2{49^\circ}} \approx 8.4 \, N$

Answer: 8.4 N