An object is hung by a thread 1.25 m long. When the object is given a horizontal velocity of 7√2 m / s it rotates in a vertical circular path What is the velocity of the object at the highest point?
Answers
Answer:
There are two forces which are acting on the body tension on the string and the weight mg
T=mv
2
+mgcosθ
By substituting the velocity as follows,
v
2
=u
2
−2ag(1−cosθ)
We get the tension as
T=
a−mg(2−3cosθ)
mu
2
At θ=60
∘
the tension in the string becomes negligible.
Hence we get
u=
2
ag
The tension at the initial point of the projection is at θ=0
∘
T=
2
3mg
Explanation:
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Answer:
There are two forces which are acting on the body tension on the string and the weight mg
T= mv
2
t mg cos
By substituting the velocity as follows,
v
2
=v
2
-2ag (1-cos)
we get the tension as
T=
a-nag (2-3cos)
mu
2
At =60
°
the tension in the string becomes negligible
Hence we get
U=
2
ag
The tension at the initial point of the projating is at=0
T=
2
3mg
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