Question

A rod of length 70.0 cm and mass 1.80 kg is suspended by two strings which are 35.0 cm long, one at each end of the rod.
$prob13a_1016full.gif$
The string on side B is cut. Find the magnitude of the initial acceleration of end B.

(in m/s^2)

A: 9.42

B: 1.18×101

C: 1.47×101

D: 1.84×101

E: 2.30×101

F: 2.87×101

G: 3.59×101

H: 4.49×101

The string on side B is retied and now has only half the length of the string on side A.
$prob13b_1016half.gif$
Find the magnitude of the initial acceleration of the end B when the string is cut.
(in m/s^2)

A: 6.78

B: 9.83

C: 1.42×101

D: 2.07×101

E: 3.00×101

F: 4.34×101

G: 6.30×101

H: 9.13×101

Answer 1

Torque after string is cut
= 1.8 g * 0.7 / 2
= 6.18 Nm
= M of I * angular accleration
= 1.8* 0.7^2 / 3 * A
= 0.294 A

0.294 A = 6.18
A = 21.02 rad /s^2

the magnitude of the initial acceleration of end B
= 21.02 * 0.7
= 14.71 m/s^2 ---answer

So here the answer is C) 1.47*10^1

angle made with vertical
= cos (-1) 0.175 / 0.7
= 75.52 deg
Torque after string is cut
= 1.8 g * 0.7 sin 75.52 / 2
= 5.98 Nm
= M of I * angular accleration
= 1.8 * 0.7^2 / 3 * A
= 0.294 A
0.294 A = 5.98
A = 20.35 rad /s^2
magnitude of the initial acceleration of the end B when the string is again cut

= 20.35 * 0.7
= 14.24 m/s^2 ---answer

So here the answer is C) 1.42*10^1