Physics Problem: Find an Algebraic Expression?
A very slippery ice cube slides in a vertical plane around the inside of a smooth, 20 cm-diameter horizontal pipe. The ice cube’s speed at the bottom of the circle is 3.0 m/s.
Find an algebraic expression for the ice cube’s speed when it is at angle θ, where the angle is measured counterclockwise from the bottom of the circle. Your expression should give 3.0 m/s for θ = 0° and 2.3m/s for θ = 180°.
Express your answer in terms of θ
So far I've tried the following equation:
sqrt [ {v_0}^2 + 2rg (cosθ} - 1) ]
where v_0 is initial velocity, r is radius of the pipe, and g is gravity, but it's the wrong answer. Apparently the answer is not supposed to contain the variable v_0 (initial velocity) or r (radius)
Any help is appreciated.
Answers & Comments
Verified answer
The good news: You're almost there!
Just substitute into your equation the values you've been given:
(1) V(θ) = sqrt ( V(0)^2 + 2 * r * g * (cos(θ) - 1) )
= sqrt ( 3^2 + 2 * 0.10 * 9.8 * (cos(θ) - 1) )
= sqrt ( 9 + 1.82 * (cos(θ) - 1) )
Substitute 0° and 180° into (1) to verify that it's correct.
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Are you OK with how to derive (1)? If not, here it is:
Since the kinetic energy KE at any point in the path of the ice cube is a function of velocity V, if you create an expression for KE as a function of θ, then you can solve it for V to get the desired expression.
The KE is at its max at the bottom, or when θ = 0; and it is at a minimum when θ = 180. As the cube ascends, its KE is reduced by its gain in potential energy PE. You can also see, since:
(11) PE = m * g * h, where h is the height of the cube,
that PE is a function of θ.
So we can write this as follows:
(12) KE(θ) = KE(0) - PE(θ)
With a little trigonometry, you can develop the following expression for h:
(13) h = r * (1 - cos(θ) )
Using the fact that KE = ½ * m * V², plus (13) and (11), we can rewrite (12) as:
(14) ½ * m * V(θ)² = ½ * m * V(0)² - m * g * r * (1 - cos(θ) )
Solving for V(θ):
(15) V(θ) = √( V(0)² - g * r * (1 - cos(θ) ) )
Or, equivalently,
(15') V(θ) = √( V(0)² + g * r * (cos(θ) - 1) ),
which is the same as your equation.
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