As shown in the figure, a uniform disc of mass m and radius r rolls along a horizontal surface and then moves up along a curved inclined surface. If the disc has a linear velocity v on the horizontal surface, and expression for moment of inertia of the disc is 𝑚𝑟2 2 , what is the maximum height the disc (its center) will travel up (Consider the possibility of the disc having a rotational kinetic energy = 1/2Iω2)? A) 𝑣2 2𝑔 B) 3𝑣2 2𝑔 C) 3𝑣2 4𝑔 D) 𝑣2 𝑔 E) 2𝑣2 g
Question
As shown in the figure, a uniform disc of mass m and radius r rolls along a horizontal surface and then moves up along a curved inclined surface. If the disc has a linear velocity v on the horizontal surface, and expression for moment of inertia of the disc is 𝑚𝑟2 2 , what is the maximum height the disc (its center) will travel up (Consider the possibility of the disc having a rotational kinetic energy = 1/2Iω2)?
A) 𝑣2 2𝑔
B) 3𝑣2 2𝑔
C) 3𝑣2 4𝑔
D) 𝑣2 𝑔
E) 2𝑣2 g
Solution
The disc has both translational and rotational kinetic energy on the horizontal surface. When it moves up the inclined surface, this kinetic energy is converted into potential energy. The maximum height is reached when all the kinetic energy has been converted into potential energy.
The total kinetic energy (K.E) of the disc on the horizontal surface is the sum of translational and rotational kinetic energy.
The translational kinetic energy (T) is given by (1/2)mv^2.
The rotational kinetic energy (R) is given by (1/2)Iω^2. But we know that I = mr^2/2 for a disc and ω = v/r (angular velocity = linear velocity/radius). Substituting these values, we get R = (1/2)(mr^2/2)(v/r)^2 = (1/4)mv^2.
So, the total K.E = T + R = (1/2)mv^2 + (1/4)mv^2 = (3/4)mv^2.
At the maximum height (h), all this kinetic energy is converted into potential energy (P.E), which is given by mgh.
Setting K.E = P.E, we get (3/4)mv^2 = mgh.
Solving for h, we get h = (3/4)(v^2/g) = 3v^2 / 4g.
So, the correct answer is C) 3v^2 / 4g.
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