Chapter 8 T=
2
mv m8gl = 8 mg R l
b) Let the velocity at C be V3 1/2 mv12 = 1/2 mv32 + mg (2l) 1/2 m (log l) = 1/2 mv32 + 2mgl v32 = 6 mgl So, the tension in the string is given by Tc =
6 glm mv 2 mg = l l
mg = 5 mg
c) Let the velocity at point D be v4 Again, 1/2 mv12 = 1/2 mv42 + mgh 1/2 × m × (10 gl) = 1.2 mv42 + mgl (1 + cos 60°) v42 = 7 gl So, the tension in the string is TD = (mv2/l) – mg cos 60° = m(7 gl)/l – l – 0.5 mg 7 mg – 0.5 mg = 6.5 mg. 54. From the figure, cos = AC/AB AC = AB cos (0.5) × (0.8) = 0.4. So, CD = (0.5) – (0.4) = (0.1) m Energy at D = energy at B 1/2 mv2 = mg (CD) v2 = 2 × 10 × (0.1) = 2 So, the tension is given by,
A
T D
mg
37°
0.5m
mv 2/ r B
0.5m
C
55. Given, N = mg As shown in the figure, mv2 / R = mg v2 = gR …(1) Total energy at point A = energy at P
k
mgR 2mgR 2
N
1/2 kx2 =
R P
m A
[because v2 = gR]
0.1 kg
mv 2 2 mg = (0.1) 10 = 1.4 N. T= r 0 .5
2
mv /R
x2 = 3mgR/k x = (3mgR ) / k . 56. V = 3gl 2
v
2
1/2 mv – 1/2 mu = –mgh v2 = u2 – 2g(l + lcos) v2 = 3gl – 2gl (1 + cos ) …(1) Again, mv2/l = mg cos v2 = lg cos From equation (1) and (2), we get 3gl – 2gl – 2gl cos = gl cos
l
m
l
u=3 gl
mv2/R
T=0
8.11
mg