For the figure shown, the
wedge + particle is resting
at a vertical wall. The
(М
particle of mass 2 is
released from rest from
the position shown. The
wedge has mass M. The
maximum height reached
by the particle on the
2.
circular groove is a R. Find
the value of a
M/2
R.
Answers
Answer:
As there is no net external force acting on the system in horizontal direction , momentum is conserved in this direction and equal to zero .
Hence , displacement of center of mass is zero .
Displacement of center of mass is given by
(S
x
)
cm
=
m
1
+m
2
m
1
S
x
1
+m
2
S
x
2
0=
m
1
+m
2
m
1
x+m
2
(x−2R)
Hence , x=
3
2R
As the mean position is taken as zero displacement and the maximum displacement is
3
2R
, the wedge is in an oscillatory motion of amplitude
3
R
about its mean position.
6mg will be the final value
Explanation:
Let a small body of mass 'm' attached to a string and revolved in a vertical circle of radius 'r'. We know that weight of the body always acts vertically downward and tension in the string towards the centre of the circular path.
Let v2 be the speed of the body and T2 be the tension in the string at the lowest point B. So at the lowest point
T2= mv2^2/r +mg
Total energy at the bottom
=PE+KE=0+ (1/2)mv^2
Let v1 be the speed and T 1 be the tension in the string at the highest point A.
So, T 1 = mv1^2/r −mg
Total energy at A=PE+KE=2mgr+ (1/2) mv1^2
Therefore
T 2 −T 1 = (m/r)(v2^2 - v1^2)+2mg
By the law of conservation of energy
Total Energy at A= Total energy at B.
Therefore
v2^2−v1^2 =4rg
Therefore
T 2 −T 1 = 4mg+2mg =6mg