Question

A mass hangs on the end of a massless rope. The pendulum is held horizontal and released from rest. When the mass reaches the bottom of its path it is moving at a speed v = 2.9 m/s and the tension in the rope is T = 14.2 N. A peg is placed 4/5 of the way down the pendulum’s path so that when the mass falls to its vertical position it hits and wraps around the peg. What is the tension in the string at the same vertical height as the peg (directly to the right of the peg)?

Answer 1

Answer:

$T = 37.5 N$

Explanation:

As the pendulum reached to the lowest position then we will have

$T - mg = \frac{mv^2}{L}$

$14.2 - m(9.81) = \frac{m(2.9^2)}{L}$

now when it will reach to the height of the peg then its speed is given as

$v_f^2 - v_i^2 = 2 a d$

so we will have

$v_f^2 - 2.9^2 = 2(-9.81)(\frac{L}{5})$

$v_f^2 = 2.9^2 - 3.924L$

also we know that

$2.9^2 - 0 = 2(9.81)(L)$

$L = 0.43 m$

$m = 0.48 kg$

now we have speed of the pendulum when it reach the same height is given as

$v_f^2 = 2.9^2 - (3.924(0.43)$

$v_f = 2.6 m/s$

Now the tension in the string is given as

$T = \frac{mv_f^2}{\frac{L}{5}}$

$T = \frac{0.48(2.6)^2}{\frac{0.43}{5}}$

$T = 37.5 N$