A small ring of mass m is attached at an end of a light string, the other end of which is tied to a small block B of mass 2m. The ring is free to move on a fixed smooth horizontal rod. Find the velocity of the ring when the string becomes vertical.?
Answers
Given:
A small ring of mass m is attached at an end of a light string, the other end of which is tied to a small block B of mass 2m.
The ring is free to move on a fixed smooth horizontal rod.
To find:
Find the velocity of the ring when the string becomes vertical.?
Solution:
From given, we have,
The mass of small ring = m
The mass of small block = 2m
The distance between the ring and block = l
We use the conservation of energy formula,
1/2 mv² = mgl + 1/3 mgl
1/2 mv² = 4/3 mgl
⇒ v² = 8/3 gl
∴ v = √(8/3 gl)
Hence the velocity of the ring when the string becomes vertical.
Answer:
Explanation:
As no external force is applied on the ring, string, mass system in x-direction.
So, conserving momentum in x-direction
⇒0=mv 1 −2mV^2⇒v 1 =2V2 ........(i)
Taking datum at rod where potential energy is zero.
⇒0= 21 mv 1 2 + 2 1 (2m)v 2 2 −2mgl ⇒2mgl= 2 1 mv 1 2 +mv 2 2
=2 1 mv 1 2 +m 4 v 12
(as v 2 = 2 v 1 from (i))
2mgl= 4 3 mv 1
2
⇒v
1
=
3
8
gl
⇒v
2
=
2
v
1
=
3
2
gl
Now relative velocity of mass with respect to ring in final condition
v
m
1
r
=v
m
1
g
−v
r
1
g
=−v
2
−v
1
=−
3
2
gl
−
3
8
gl
∣v
m
1
r
∣=
gl
(
3
2
+
8
).
Now drawing FBD of mass in final condition.
⇒T−2mg=2m
l
(v
m
1
r
)
2
⇒T−2mg=2m
l
gl
(
3
2
+
8
)
2
⇒T=2mg
⎝
⎛
1+(
3
2
+
8
)
2
⎠
⎞
=14mg
=14×
2
1
×10 (Taking g=10m/s
2
)
T=70N.
solution