# Answer to Question #5753 in Mechanics | Relativity for Melanie

Question #5753

A string under a tension of 43 N is used to whirl a rock in a horizontal circle of radius 3.3 m at a speed of 26.46 m/s. The string is pulled in, and the speed of the rock increases. When the string is 0.635 m long and the speed of the rock is 80.5 m/s, the string breaks. What is the breaking strength of the string? Answer in units of N.

Expert's answer

Let's make the following denominations:

F1 = 43 N

R1 = 3.3 m

V1 = 26.46 m/s

R2 = 0.635 m

V2 = 80.5 m/s

M - the mass of a rock

We need to find the mass of a rock at first.

Using the first case we have that the centripetal acceleration of a rock is

Ac1 = V1²/R1 = 26.46²/3.3 ≈ 212.1611 m/s².

Now let's use the second Newton's law:

F1 = M*Ac1 ==> M = F1/Ac1 = 43/212.1611 = 0.2027 kg.

Now let's find the breaking strength of the string.

Using the second case we have that the centripetal acceleration of a rock is

Ac2 = V2²/R2 = 80.5²/0.635 ≈ 10205.1181 m/s².

So, the breaking strength is

F2 = M*Ac2 = 0.2027*10205.1181 = 2068.5774 N.

F1 = 43 N

R1 = 3.3 m

V1 = 26.46 m/s

R2 = 0.635 m

V2 = 80.5 m/s

M - the mass of a rock

We need to find the mass of a rock at first.

Using the first case we have that the centripetal acceleration of a rock is

Ac1 = V1²/R1 = 26.46²/3.3 ≈ 212.1611 m/s².

Now let's use the second Newton's law:

F1 = M*Ac1 ==> M = F1/Ac1 = 43/212.1611 = 0.2027 kg.

Now let's find the breaking strength of the string.

Using the second case we have that the centripetal acceleration of a rock is

Ac2 = V2²/R2 = 80.5²/0.635 ≈ 10205.1181 m/s².

So, the breaking strength is

F2 = M*Ac2 = 0.2027*10205.1181 = 2068.5774 N.

## Comments

## Leave a comment