Young David, who slew Goliath, experi- mented with slings before tackling the gi- ant. He found that with a sling of length 0.566 m, he could revolve the sling at the rate of 10.7 rev/s. If he increased the length to 0.87 m, he could revolve the sling only 7.7 rev/s.

Question

Young David, who slew Goliath, experi- mented with slings before tackling the gi- ant. He found that with a sling of length 0.566 m, he could revolve the sling at the rate of 10.7 rev/s. If he increased the length to 0.87 m, he could revolve the sling only 7.7 rev/s.

a) What is the larger of the two linear speeds?
Answer in units of m/s.

part 2

b) Using the sling length 0.87 m, what is the centripetal acceleration at 7.7 rev/s?
Answer in units of m/s2.

Answer 1

To answer these questions, we need to apply the concepts of linear speed and centripetal acceleration. Let's break down the steps to find the answers.

a) What is the larger of the two linear speeds?

Linear speed is given by the formula:
Linear Speed = 2πr/T

where:
- Linear Speed is the speed at which an object moves along a circular path
- r is the radius of the circular path
- T is the time it takes to complete one revolution (in this case, it is given in rev/s)

Given the lengths of the two slings, we can determine the radii as follows:
- For the sling with a length of 0.566 m, the radius is half the length, i.e., r₁ = 0.566 m / 2 = 0.283 m.
- For the sling with a length of 0.87 m, the radius is half the length, i.e., r₂ = 0.87 m / 2 = 0.435 m.

Now, we can calculate the linear speeds associated with each radius using the given revolution rates:
- For the first case, r₁ = 0.283 m and T₁ = 10.7 rev/s
- For the second case, r₂ = 0.435 m and T₂ = 7.7 rev/s

Using the formula given above, we can calculate the linear speeds:
Linear Speed₁ = 2π(0.283) / 10.7
Linear Speed₂ = 2π(0.435) / 7.7

Comparing the two linear speeds, the larger one is the answer to part a) and would be in units of m/s.

b) Using the sling length 0.87 m, what is the centripetal acceleration at 7.7 rev/s?

Centripetal acceleration is given by the formula:
Centripetal Acceleration = (Linear Speed)² / r

To solve this question, we already have the linear speed (from part a)) and the radius:
- Linear Speed = Linear Speed₂ (from part a) calculation)
- r = 0.435 m

Plugging in the values into the formula, we can calculate the centripetal acceleration:
Centripetal Acceleration = (Linear Speed)² / r

The answer to part b) would be the calculated centripetal acceleration in units of m/s².