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I have solved the following sum but when I am calculating the result the answers seems different although I am sure about the method. ?
The answer is:
(12⋅m⋅(v0)2)−(12⋅m⋅g⋅sin(ϕ)⋅R)−(μ⋅m⋅g⋅cos(ϕ)⋅π)−(2⋅m⋅g⋅sin(ϕ)⋅R)R
Where am I doing wrong exactly?
Q) A body of mass m is attached to one end of a string of length R. The other end of the string is fixed on a inclined surface making an angle ϕ with the surface as shown in the figure. The body has a speed v0 at the bottom of the circle (Point A). The body undergoes circular motion. There is a coefficient of sliding friction μ between the body and the plane. The downward acceleration of gravity is g. what is the tension in the string when it reaches point B?
1 Answer
- ?Lv 72 weeks agoFavorite Answer
Without showing all of your work, it's difficult to know where errors occurred. Here's how I would do it.
If we set A as the gravitational reference plane.
The kinetic energy at B will equal the kinetic energy at A minus the potential energy at B minus the work of friction between A and B.
½mv₁² = ½mv₀² - mgh - Fd
½mv₁² = ½mv₀² - mg(2Rsinφ) - (μmgcosφ)(πR)
v₁² = v₀² - 4Rgsinφ - 2μπRgcosφ
The tension in the string at B will need to supply the necessary centripetal acceleration minus the contribution of gravity
T = m(v₁²/R - gsinφ)
T = m((v₀² - 4Rgsinφ - 2μπRgcosφ)/R - gsinφ)
T = m(v₀²/R - 4gsinφ - 2μπgcosφ - gsinφ)
T = m(v₀²/R - 5gsinφ - 2μπgcosφ)
T = m(v₀²/R - g(5sinφ + 2μπcosφ))
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