Consider a point mass m attached to the end of a massless string. The other end

Question

Consider a point mass m attached to the end of a massless string. The other end of the massless rod is fixed to a wall as shown in the figure. Let ? be the angle that the rod makes with the vertical. Assume that the mass is pulled to one side such that ? = ? 2 . Then it is released with zero initial speed. A m ? (a) Using conservation of mechanical energy, calculate the kinetic energy and the speed of the mass m when it passes through the bottom point (? = 0) (b) What is the moment of inertia of the mass m relative to rotations around the point A? (c) What is the angular velocity of the mass m as it goes through the bottom point? (d) What is the rotational energy of the mass m for rotations around the point A as it goes through the bottom point? (e) Compare the rotational energy that you calculated in the previous part with the kinetic energy that you calculated in part (a)

Explanation / Answer

a point mass m attached to the end of a massless rod

Length of rod = L

= /2

Energy conservation

(a.)Initial potential energy = final kinetic energy

M*g*L = final kinetic energy

Kinetic energy at bottom = MgL

1/2 MV2 = MgL

V =(2gL)1/2

(b.) Moment of inertia of mass about A = ML2

(c.) Angular velocity = V/L = (2g/L)1/2

(d.) Rotational energy = 1/2 Iw2   where I = moment of inertia and w = angular velocity

= 1/2 ML2 *2g/L = MgL

(e.) Both are same

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