Motion in a vertical circle and conservation of energy
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?
Answer:
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
[latex]\frac{1}{2}mu^{2}=\frac{1}{2}mv^{2}+mgl[/latex]
From the equation above, v-u can be calculated.
[latex]v=\sqrt{u^{2}-2gl}[/latex]
The following links will help you for deeper understanding and you can browse through some solved problems from the topic too.
Categories: Answers, ANSWERS TO QUESTIONS FROM VISITORS, CLASS XI, energy loss, energy motion, Exam Help, General Physics, IIT JEE, Interesting Questions, motion work, Physics Homework, Plus Two Physics, power and energy, Problem Solving, Problems, Soved Numerical Problems Tags: Auto, centre, conservation of energy, law of conservation, law of conservation of energy, magnitude, motion, motion laws, position, stone, vertical circle, vertical height
A Numerical Problem from Projectile Motion
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)
Answer:
Take Sy= -90m
ay= -g=-9.8m/s2
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
Categories: Answers, Ask Physics, Mechanics, Problem Solving, Problems, Project Tags: answer, hill, horizontal velocity, motion, numerical, numerical problem, ponit, Projectile motion, Substitute, x time
An MCQ from Projectile motion
Devender posted this question: A plane is flying horizontally at 98m/s and releases an object, which reaches the ground in 10s. the angle made by it while hitting the ground is
(a) 55 degree
(b) 45 degree
(c) 60 degree
(d) 75 degree
Categories: KINEMATICS, Problem Solving, Unsolved Numerical Problems Tags: degree, degree c, devender, ground, hitting the ground, MCQ, plane, Projectile, Projectile motion
A bullet of mass 20g enters into a fixed block with aspeed 40m/s and stops in it. find the change in internal energy during the process.
A bullet of mass 20 g enters into a fixed block with a speed 40m/s and stops in it.
find the change in internal energy during the process?
Harsh Asked.
The question is open to visitors to post answer as comment
Categories: +2 Physics, Answers, Problem Solving Tags: bullet, energy, internal energy, work energy theorem
+2 Physics :: Problem Solving Tactics in Physics
Physics is often refered to as a Problem Solving Discipline. One who has acquired the skills of problem solving will certainly excel in Physics. The students who learn +2 Physics mainly complain on the difficulties in Problem Solving. It is very essential to get trained in solving problems in Physics from as many sources as possible.
Please refer to the following links for more exposure on various problem solving techniques and strategies as well as tactics in solving Physics Problems.
http://faculty.normandale.edu/~physics/Hollabaugh/probsolv.htm
