A stone tied to a string of length l is whirled around a vertical circle with the other end of the string at the centre. At a certain instant of time the stone is at the lowest position and has a speed u. What is the magnitude of change in its velocity as it reaches a position where the string is horizontal?
Let’s assume that the potential energy at the lowest position be zero. So, when the string is horizontal, the stone has risen by a vertical height l, the length of the string which is also the radius of the vertical circle.
If v is the magnitude of velocity at the horizontal position, then according to the law of conservation of energy,
KE+PE at the lowest position = KE+PE at the horizontal position
From the equation above, v-u can be calculated.
The following links will help you for deeper understanding and you can browse through some solved problems from the topic too.
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A light year is the distance light travels in one year . How many meters are there in one light year?
The speed of light in vacuum is 3 x 10^8 m/s. That means in one second it travels 300000000 metres. As we know, in one year, there are 365 days; in each day there are 4 hours; in each hour there are 60 minutes and in each minute there are 60 seconds. Therefore, the distance [...]
A short is fired from a gun on the top of a hill 90m high with velocity 80m/s at an angle of 30 degree with the horizontal. Find the horizontal distance between the vertical line through the gun and the ponit where the shot strikes the ground? (Shilpa asked this question)
Take Sy= -90m
uy= u sin(30) =80 x 0.5 = 40m/s
Substitute in the eqn
s = ut + 0.5 at2
and find t
The horizontal distance Rx = horizontal velocity x time = U cos (30) x time
Substitute for u and t and get the answer
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Natasha Sehgal Asked:
“A stone is thrown in a vertically upward direction with a velocity of 5m/s. If the acceleration of the stone during its motion is 10m/s^2 in the downward direction , what will be the height attained by the stone and how much time will it take to reach there?”
a= -10 m/s2
v=0 at the topmost position
S= Hmax = ?
The eqn of motion to use
v2 – u2 = 2aS
Which gives H = 1.25 m
From the eqn
v=u + at
t= 0.5 s