Worked examples · Sheet 2

Flywheels and pulleys with mass

Theory used: Angular position, velocity and acceleration Velocity and acceleration of the points of the body Torque of a force Moments of inertia of uniform bodies Equation of rotational motion about a fixed axis

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

A flywheel in the shape of a uniform disk, of mass $M = 20 kg$ and radius $R = 0.30 m$ , rotates about its own axis at $300 rpm$ . A brake shoe presses on the rim and exerts a constant friction force tangent to the rim, of magnitude $F = 15 N$ . Neglecting the friction of the axle, compute the angular acceleration, the stopping time and the number of revolutions made during braking.

I = \frac{1}{2} M R^{2} = \frac{1}{2} \cdot 20 \cdot 0.3 0^{2} = 0.90 kg \cdot m^{2}, τ_{z} = - F R = - 15 \cdot 0.30 = - 4.5 N \cdot m

τ_{z} = I α ⟹ α = \frac{τ _{z}}{I} = \frac{- 4.5}{0.90} = - 5.0 rad/ s^{2}

ω_{0} = 300 \cdot \frac{2 π}{60} rad/s = 10 π rad/s = 31.4 rad/s

0 = ω_{0} + α t ⟹ t = \frac{ω _{0}}{∣ α ∣} = \frac{31.4}{5.0} = 6.28 s

0 = ω_{0}^{2} + 2 α Δ θ ⟹ Δ θ = \frac{ω _{0}^{2}}{2∣ α ∣} = \frac{( 10 π ) ^{2}}{10} = 98.7 rad, n = \frac{Δ θ}{2 π} = 15.7 revolutions

The friction force is tangent to the rim, so its lever arm about the axis is the radius, and the torque is negative because it opposes the rotation. The flywheel rotates about a frictionless fixed axis, therefore the equation of rotational motion $τ_{z} = I α$ applies, with the moment of inertia of the disk. The torque is constant, so $α$ is also constant and the laws of uniformly accelerated rotation hold. The initial angular velocity is converted to $rad/s$ by multiplying the revolutions by $2 π$ and dividing the minutes by $60$ . The stopping time is obtained by setting $ω = 0$ , the angle described from the relation between $ω^{2}$ and $Δ θ$ ; dividing the angle by $2 π$ we obtain the number of revolutions, that is $15.7$ revolutions in $6.28 s$ .

Exercise 2

A uniform cylinder of mass $M = 6.0 kg$ and radius $R = 0.15 m$ rotates without friction about its own horizontal axis. An inextensible string of negligible mass is wound around it; the string does not slip and supports a block of mass $m = 2.0 kg$ . The system starts from rest. Compute the acceleration of the block, the tension of the string, the angular acceleration of the cylinder and the time the block takes to descend by $h = 1.5 m$ .

⎩ ⎨ ⎧ m g - T = ma T R = \frac{1}{2} M R^{2} α a = α R

T = \frac{1}{2} M a, m g - \frac{1}{2} M a = ma ⟹ a = \frac{m g}{m + \frac{M}{2}}

a = \frac{2.0 \cdot 9.81}{2.0 + 3.0} m/ s^{2} = 3.92 m/ s^{2}, T = 3.0 \cdot 3.924 N = 11.8 N

α = \frac{a}{R} = \frac{3.924}{0.15} rad/ s^{2} = 26.2 rad/ s^{2}

h = \frac{1}{2} a t^{2} ⟹ t = \frac{2 h}{a} = \frac{2 \cdot 1.5}{3.924} s = 0.874 s

The block translates and the cylinder rotates, so we write Newton’s second law for the block, taken positive downward, and the equation of rotational motion for the cylinder. On the cylinder the tension acts tangent to the rim with lever arm $R$ , while the weight and the reaction of the axle have zero torque, because they are applied on the axis. The string is inextensible and does not slip, therefore the acceleration of the block is equal to the tangential acceleration of the rim. From the last two equations we obtain $T$ as a function of $a$ , and substituting into the first we obtain $a$ . The motion of the block is uniformly accelerated from rest, which gives the descent time. As a check, for $M \to 0$ the formula gives $a \to g$ , that is free fall; with the cylinder the tension $11.8 N$ is smaller than the weight of the block, $19.6 N$ , and the block descends with $a = 3.92 m/ s^{2}$ .

Exercise 3

An Atwood machine consists of two blocks of masses $m_{1} = 3.0 kg$ and $m_{2} = 2.0 kg$ , connected by an inextensible string of negligible mass that passes over a pulley in the shape of a uniform disk of mass $M = 2.0 kg$ and radius $R = 0.10 m$ . The pulley rotates without friction and the string does not slip on it. Compute the acceleration of the blocks and the tensions in the two segments of string.

Forces on the two blocks: the tensions

T_{1}

and

T_{2}

are different because the pulley has mass

⎩ ⎨ ⎧ m_{1} g - T_{1} = m_{1} a T_{2} - m_{2} g = m_{2} a (T_{1} - T_{2}) R = \frac{1}{2} M R^{2} α a = α R

T_{1} - T_{2} = \frac{1}{2} M a, (m_{1} - m_{2}) g - \frac{1}{2} M a = (m_{1} + m_{2}) a ⟹ a = \frac{( m _{1} - m _{2} ) g}{m _{1} + m _{2} + \frac{M}{2}}

a = \frac{1.0 \cdot 9.81}{3.0 + 2.0 + 1.0} m/ s^{2} = 1.64 m/ s^{2}

T_{1} = m_{1} (g - a) = 3.0 \cdot (9.81 - 1.635) N = 24.5 N, T_{2} = m_{2} (g + a) = 2.0 \cdot (9.81 + 1.635) N = 22.9 N

The heavier block $m_{1}$ descends and $m_{2}$ rises with the same acceleration $a$ , because the string is inextensible; for each block Newton’s second law is written with the positive sense in the direction of motion. On the pulley the two segments of string pull downward, tangent to the rim on opposite sides, so their torques about the axis have opposite signs and the resultant torque is $(T_{1} - T_{2}) R$ . The equation of rotational motion of the pulley, with $I = \frac{1}{2} M R^{2}$ , and the no-slip condition give $T_{1} - T_{2} = \frac{1}{2} M a$ . Adding the first two equations, the tensions appear only in the difference $T_{1} - T_{2}$ , and $a$ is obtained. As a check, $T_{1} - T_{2} = 1.6 N$ coincides with $\frac{1}{2} M a = 1.0 \cdot 1.635 N$ . The pulley with mass behaves like an additional mass $\frac{M}{2}$ to be accelerated, therefore $a = 1.64 m/ s^{2}$ is smaller than the value $1.96 m/ s^{2}$ that would be obtained with a pulley of negligible mass.

Exercise 4

A block of mass $m_{1} = 4.0 kg$ rests on a horizontal table with coefficient of kinetic friction $μ_{d} = 0.20$ and is connected, by an inextensible string of negligible mass that passes over a pulley at the edge of the table, to a hanging block of mass $m_{2} = 3.0 kg$ . The pulley is a uniform disk of mass $M = 2.0 kg$ and radius $R = 0.080 m$ ; it rotates without friction and the string does not slip. The system, released, starts to move. Compute the acceleration and the tensions in the two segments of string.

⎩ ⎨ ⎧ T_{1} - μ_{d} m_{1} g = m_{1} a m_{2} g - T_{2} = m_{2} a (T_{2} - T_{1}) R = \frac{1}{2} M R^{2} α a = α R

a = \frac{( m _{2} - μ _{d} m _{1} ) g}{m _{1} + m _{2} + \frac{M}{2}} = \frac{( 3.0 - 0.80 ) \cdot 9.81}{4.0 + 3.0 + 1.0} m/ s^{2} = 2.70 m/ s^{2}

T_{1} = m_{1} (a + μ_{d} g) = 4.0 \cdot (2.698 + 1.962) N = 18.6 N

T_{2} = m_{2} (g - a) = 3.0 \cdot (9.81 - 2.698) N = 21.3 N

The block on the table is pulled by the tension $T_{1}$ and slowed by kinetic friction, which equals $μ_{d} m_{1} g$ because the normal reaction of the table balances the weight. The hanging block descends under the action of the weight and of the tension $T_{2}$ . On the pulley the two segments of string act tangent to the rim, one horizontal and one vertical, with opposite torques about the axis; the equation of rotational motion applies with $I = \frac{1}{2} M R^{2}$ . Adding the first two equations and using $T_{2} - T_{1} = \frac{1}{2} M a$ , which follows from the last two, we obtain $a$ . The radius of the pulley cancels and does not appear in the result. The check $T_{2} - T_{1} = 2.7 N = \frac{1}{2} M a$ confirms the values, and the tension is greater in the vertical segment, because the difference of the tensions accelerates the rotation of the pulley.