Worked examples · Sheet 3

Conservation of energy in rotational motion

Theory used: Rotational kinetic energy Gravitational potential energy of a rigid body Conservation of mechanical energy Work and power in rotations Moments of inertia of uniform bodies

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

A uniform disk of mass $M = 4.0 kg$ and radius $R = 0.20 m$ rotates without friction about its own horizontal axis. An inextensible string of negligible mass wound around the rim supports a block of mass $m = 1.0 kg$ ; the string does not slip. The system starts from rest. Using conservation of energy, compute the velocity of the block and the angular velocity of the disk when the block has descended by $h = 2.0 m$ .

m g h = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}, I = \frac{1}{2} M R^{2}, ω = \frac{v}{R}

m g h = \frac{1}{2} m v^{2} + \frac{1}{2} \cdot \frac{1}{2} M R^{2} \cdot \frac{v ^{2}}{R ^{2}} = \frac{1}{2} (m + \frac{M}{2}) v^{2}

v = \frac{2 m g h}{m + \frac{M}{2}} = \frac{2 \cdot 1.0 \cdot 9.81 \cdot 2.0}{1.0 + 2.0} m/s = 3.62 m/s

ω = \frac{v}{R} = \frac{3.617}{0.20} rad/s = 18.1 rad/s

The axle is frictionless and the string does not slip, so the reactions of the constraints do no work and the only force that does work is the weight of the block; by the conservation of mechanical energy the potential energy lost by the block becomes kinetic energy of the block and of the disk. The disk rotates about a fixed axis, therefore its kinetic energy is $\frac{1}{2} I ω^{2}$ , and the speed of the block is equal to the speed of the rim, $v = ω R$ . The radius cancels, and the disk contributes like an additional mass $\frac{M}{2}$ . The block arrives with $v = 3.62 m/s$ . As a check, the dynamics gives $a = \frac{m g}{m + \frac{M}{2}} = 3.27 m/ s^{2}$ and $v^{2} = 2 ah = 13.1 m^{2} / s^{2}$ , in agreement with the result.

Exercise 2

A uniform rod of length $L = 1.5 m$ is hinged without friction at one end $O$ and can swing in a vertical plane. It is released from rest when it forms an angle of $6 0^{\circ}$ with the downward vertical. Compute the angular velocity of the rod and the speed of the free end when the rod passes through the vertical.

From the initial position to the vertical one, dashed, the center of mass descends by

Δ h = \frac{L}{2} (1 - cos 6 0^{\circ}) = \frac{L}{4}

Δ h = \frac{L}{2} (1 - cos 6 0^{\circ}) = \frac{L}{4} = 0.375 m

M g \frac{L}{4} = \frac{1}{2} \cdot \frac{1}{3} M L^{2} ω^{2} ⟹ ω = \frac{3 g}{2 L}

ω = \frac{3 \cdot 9.81}{2 \cdot 1.5} rad/s = 3.13 rad/s, v = ω L = 3.132 \cdot 1.5 m/s = 4.70 m/s

The hinge is frictionless, so its reaction does no work, and the weight is the only force that does work; therefore mechanical energy is conserved. For the potential energy of a rigid body only the height of the center of mass matters, and the center of mass is at the midpoint of the rod. In the initial position the center of mass is $\frac{L}{2} cos 6 0^{\circ}$ below the hinge, in the vertical position $\frac{L}{2}$ , so it descends by $\frac{L}{4}$ . The rod rotates about the fixed axis through $O$ , therefore its kinetic energy is $\frac{1}{2} I_{O} ω^{2}$ with $I_{O} = \frac{1}{3} M L^{2}$ . The mass cancels, and the speed of the end is $ω L$ because the end is at distance $L$ from the axis. In the vertical position the rod has $ω = 3.13 rad/s$ and the free end $v = 4.70 m/s$ .

Exercise 3

A flywheel in the shape of a uniform solid cylinder, of mass $50 kg$ and radius $0.40 m$ , is brought from rest to $3000 rpm$ by a motor that delivers a constant power of $500 W$ . Neglecting friction, compute the final kinetic energy of the flywheel, the time required and the torque applied by the motor at the final speed.

I = \frac{1}{2} M R^{2} = \frac{1}{2} \cdot 50 \cdot 0.4 0^{2} = 4.0 kg \cdot m^{2}, ω = 3000 \cdot \frac{2 π}{60} rad/s = 100 π rad/s = 314 rad/s

K = \frac{1}{2} I ω^{2} = \frac{1}{2} \cdot 4.0 \cdot (100 π)^{2} J = 1.97 \cdot 1 0^{5} J

W = P t = K ⟹ t = \frac{K}{P} = \frac{1.974 \cdot 1 0 ^{5}}{500} s = 395 s

P = τ_{z} ω ⟹ τ_{z} = \frac{P}{ω} = \frac{500}{314.2} N \cdot m = 1.59 N \cdot m

The rotational kinetic energy of the flywheel depends on the moment of inertia of the solid cylinder and on the angular velocity, converted to $rad/s$ . Friction is negligible, so by the work-energy theorem the work of the motor, denoted by $W$ to distinguish it from the angular momentum, is equal to the final kinetic energy. The power is constant, therefore the work is the product of the power and the time, and the flywheel reaches the final speed in $395 s$ , that is in about $6.6$ minutes. From the relation $P = τ_{z} ω$ it follows that, at constant power, the torque of the motor decreases as the angular velocity increases, and at the final speed it equals $1.59 N \cdot m$ .

Exercise 4

A compound pendulum consists of a uniform rod of mass $M = 1.0 kg$ and length $L = 0.60 m$ , hinged without friction at one end, and of a small ball of mass $m = 0.50 kg$ , of negligible size, fixed at the other end. The pendulum is released from rest in a horizontal position. Compute the speed of the ball when the pendulum passes through the vertical and compare it with the free-fall speed from the same height.

I_{O} = \frac{1}{3} M L^{2} + m L^{2} = \frac{1}{3} \cdot 1.0 \cdot 0.6 0^{2} + 0.50 \cdot 0.6 0^{2} = 0.12 + 0.18 = 0.30 kg \cdot m^{2}

Δ U = M g \frac{L}{2} + m g L = (0.30 + 0.30) \cdot 9.81 J = 5.89 J

\frac{1}{2} I_{O} ω^{2} = Δ U ⟹ ω = \frac{2 Δ U}{I _{O}} = \frac{2 \cdot 5.886}{0.30} rad/s = 6.26 rad/s

v = ω L = 6.264 \cdot 0.60 m/s = 3.76 m/s, v_{free} = 2 g L = 3.43 m/s

The pendulum is a rigid body rotating about a frictionless fixed axis, so mechanical energy is conserved. The moment of inertia about the axis is the sum of that of the rod about one end and that of the ball, treated as a point mass at distance $L$ . The loss of potential energy is the sum of the contributions of the two parts, since the center of mass of the rod descends by $\frac{L}{2}$ and the ball by $L$ . Equating the rotational kinetic energy to the loss of potential energy we obtain $ω$ , and the ball, which is at distance $L$ from the axis, has speed $ω L$ . The ball arrives with $3.76 m/s$ , faster than in free fall, because the rod, through the internal forces, transfers to it part of the potential energy lost by its own points.