Answer to Question #85829 in Mechanics | Relativity for Julie

Question

A 1.3 kg ball at the end of a 1.1 m string swings in a vertical plane. At its lowest point the ball is moving with a speed of 14 m/s. (a) What is its speed (in m/s) at the top of its path? (b) What is the tension (in N) in the string when the ball is at the bottom and at the top of its path?

Answer 1

Solution:

Let's write the law of conservation of energy of the ball for two positions - when ball is at the bottom and at the top of its path:

$$\frac{mv^2_{top}} 2+2mgl=\frac{mv^2_{bottom}} 2$$

From this equation we can find the speed of the ball at the top of its path:

$$v_{top}=\sqrt{v^2_{bottom}-4gl}$$

$$v_{top}=\sqrt{14^2-4\cdot9.8\cdot1.1}=\sqrt{153}=12 (m/s)$$

Let's find the tension in the string when the ball is at the top of its path. Newton's second law of motion for this case:

$$mg+T_{top}=ma_{top}=\frac{mv^2_{top}} l$$

$$T_{top}=\frac{mv^2_{top}} l-mg=m(\frac{v^2_{top}} l-g)$$

$$T_{top}=1.3(\frac{12^2} {1.1}-9.8)=157 (N)$$

Let's find the tension in the string when the ball is at the bottom of its path. Newton's second law of motion for this case:

$$T_{bottom}-mg=ma_{bottom}=\frac{mv^2_{bottom}} l$$ $$T_{bottom}=\frac{mv^2_{bottom}} l+mg=m(\frac{v^2_{bottom}} l+g)$$ $$T_{bottom}=1.3(\frac{14^2} {1.1}+9.8)=244 (N)$$

Answer: $v_{top}=12 m/s,$ $T_{top}=157 N,$ $T_{bottom}=244 N$