Question

A rope swing is hung from a tree right at the edge of a small creek. The rope is 5.0 m long; the creek is 3.0 m wide.

Answer 1

Answer:

1) T = 4.5 s

2) T = 4.5 s

3) v = 9.9 m/s

Explanation:

We can use the equation

T = 2π√(L/g)

1) T = 2π√(5m/9.81 m/s²) = 4.5 s

2) T = 2π√(L/g)

T = 2π√(5m/9.81 m/s²) = 4.5 s

3) v = √(2gR)

v = √(2(9.81 m/s²)(5m))

v = 9.9 m/s

Answer 2

Answer:

2.24

2.24

6.3 m/s

Explanation:

- The given problem can be modeled like a swinging pendulum.

- We will use already derived expressions for SHM of a simple pendulum, so for the time period we have the expression:

                                       T = 2*pi * sqrt ( L / g )

                                       T = 2*pi*sqrt (5.0/9.81)

                                       T = 4.4857 s

- The entire cycle takes 4.4857 s, to complete. However, we are asked to find the half cycle or that is the first time he reaches his starting point.

- Hence, 1/2 T                 t= 2.24 s

- We can see that the time period of the SHM is independent of the mass of the object, hence the answer to second question is also t = 2.24

c)

- In this case we will have to determine the angle that the rope makes with the vertical position:

- We are told its right across the creek, so the angle can be computed:

                                        cos(Q) = 3 / 5

- Now the maximum velocity of a pendulum is given by:

                                        v_max = sqrt(2*g*L*(1-cosQ))

- plug in the values:        v_max = sqrt(2*9.81*5*(1-3/5))

                                        v_max = 6.26 m/s

Hence option A is 6.3 m/s is correct.