Unit 8: Rotational Motion

It increases.

The rod in Figure Q8.13 is an example of an object whose rotational inertia differs depending on the rotational axis. Note that when the axis of rotation shifts from the center of the rod to one of its ends, the rotational inertia increases by four times.

The torques are equal in magnitude.

A clockwise torque tends to turn a system in the clockwise direction, while a counterclockwise torque does the opposite. A balanced system is in a state of mechanical equilibrium and experiences a net torque equal to zero. Therefore, the two opposing torques on the system must have the same magnitude.

1. $20\,\text{N}\cdot\text{m}$
2. 200 N
3. Yes.

Assume the force is exerted perpendicular to the wrench handle. When you push with a force of 80 N at a distance of 0.25 m from the axis of the bolt, you exert a torque given as \begin{align} \text{Torque} &= \text{lever arm} \times \text{force} \nonumber\\[6pt] &= 0.25\,\text{m} \times 80\,\text{N} \nonumber\\[6pt] &= 20\,\text{N}\cdot\text{m} \end{align}

When you reduce the lever arm to 0.10 m, you achieve the same torque by exerting a force $F$ calculated as follows: \begin{align} 20\,\text{N}\cdot\text{m} &= 0.10\,\text{m} \times F \nonumber\\[6pt] \Rightarrow\quad F &= \frac{20\,\text{N}\cdot\text{m}}{0.10\,\text{m}} \nonumber\\[6pt] &= 200\,\text{N} \end{align} You maximize the force applied to the lever arm when you direct it perpendicular to the handle. Therefore, the torque applied on the bolt depends on the direction of your push relative to the direction of the wrench handle.

1 kg

The mass of the 1-kg meterstick is divided into 0.25 kg for the shorter section and 0.75 kg for the longer section. To ensure equilibrium, the net torque about the 25-cm mark must be equal to zero. As shown in Figure Q8.50(b), equilibrium is achieved when the counterclockwise torque produced by the combined weight of the rock and the shorter section of the meterstick is balanced by the clockwise torque produced by the weight of the longer section.

The weight of the longer section of the meterstick is equal to 7.5 N and is represented by a downward arrow at its center of mass (i.e., 37.5 cm to the right of the fulcrum). So the clockwise torque on the system is given as

\begin{align} \text{Clockwise torque} &= 7.5\,\text{N} \times 0.375\,\text{m} \nonumber\\[6pt] &= 2.8125\,\text{N}\cdot\text{m} \label{ch08_q46_clockwise} \end{align}

Similarly, the shorter section of the meterstick has a weight of 2.5 N, which acts 12.5 cm to the left of the fulcrum. If $mg$ is the rock’s weight, the net counterclockwise torque on the system is given by

\begin{align} \text{Counterclockwise torque} &= (mg \times 0.25\,\text{m}) + (2.5\,\text{N} \times 0.125\,\text{m}) \nonumber\\[6pt] &= (mg \times 0.25\,\text{m}) + 0.3125\,\text{N}\cdot\text{m} \label{ch08_q46_counterclockwise} \end{align}

Since the system is balanced, Equations 8.15 and 8.16 are equal and you can write \begin{align} (mg \times 0.25\,\text{m}) + 0.3125\,\text{N}\cdot\text{m} &= 2.8125\,\text{N}\cdot\text{m} \nonumber\\[6pt] \Rightarrow\quad mg \times 0.25\,\text{m} &= 2.5\,\text{N}\cdot\text{m} \nonumber\\[6pt] \Rightarrow\quad mg &= 10\,\text{N} \end{align} You can then calculate the mass of the rock: \begin{align} m &= \frac{10\,\text{N}}{g} \nonumber\\[6pt] &= \frac{10\,\text{N}}{10\,\text{m}/\text{s}^2} \nonumber\\[6pt] &= 1\,\text{kg} \end{align}

1. $T = \dfrac{m v^2}{\ell}$
2. $m = \dfrac{\ell T}{v^2}$
3. $m = \dfrac{2\,\text{m} \times 10\,\text{N}}{(2\,\text{m/s})^2} = 5\,\text{kg}$

1. The centripetal force here is provided by the tension $T$ in the string, which causes the puck to rotate in a circular path of radius $\ell$. So the relevant equation in this case is expressed as \begin{equation} T = \frac{m v^2}{\ell} \end{equation}

2. When multiplying both sides of the equation above by $\ell$ and dividing by $v^2$, you can write

   \begin{align} \frac{\ell}{v^2} \times T &= \frac{\ell}{v^2} \times \frac{m v^2}{\ell} \nonumber\\[6pt] \Rightarrow\quad m &= \frac{\ell T}{v^2} \label{ch08_q51_m} \end{align}

3. You calculate the mass of the puck by substituting $\ell = 2\,\text{m}$, $T = 10\,\text{N}$, and $v = 2\,\text{m/s}$ into Equation 8.20: \begin{align} m &= \frac{(2\,\text{m}) (10\,\text{N})}{(2\,\text{m/s})^2} \nonumber\\[6pt] &= 5\,\text{kg} \end{align}

Three rotations per second.

Neglecting the external torque due to air resistance, the angular momentum ($I \omega$) of the trapeze artist is conserved. Since, her rotational inertia ($I$) reduces to one-third its initial value, the angular velocity ($\omega$) must triple in order to keep the angular momentum constant. Mathematically, this can be expressed as \begin{equation} I \omega = \dfrac{1}{3}\, I\times 3\, \omega \end{equation}

B is the winner, A comes in second place, and C comes in third place.

A round object with greater rotational inertia has more resistance to changing its rotational speed than a round object with less rotational inertia. Referring to Table 8.1 in the eText, you see that the rotational inertia per unit mass for the three objects in the race is given by \begin{align} \frac{I}{m} &= r^2 \quad (\text{hoop}) \\[6pt] \frac{I}{m} &= \dfrac{1}{2} r^2 = 0.5\, r^2 \quad (\text{solid disk}) \\[6pt] \frac{I}{m} &= \dfrac{2}{5} r^2 = 0.4\, r^2 \quad (\text{solid ball}) \end{align} Therefore, the solid ball in Figure Q8.56 has the greatest acceleration, followed by the solid disk and then the hoop.

Note that the end of the flyswatter moves over a larger arc than your hand does during the same time interval.

As shown in Figure Q8.64, your hand and the flyswatter are lined up and rotate with the same rotational speed ($\omega$) with respect to the wrist joint. Therefore, the linear speed ($v$) depends on the radius of rotation ($r$) and is given by \begin{equation} v = r \omega \end{equation} Since the radius of rotation of the flyswatter is greater than the radius of rotation of your hand, the flyswatter moves faster.

Note that the solid sphere has less rotational inertia compared with the hollow sphere.

You would release the two spheres from rest at the top of the incline to see which one reaches the bottom first. The entire mass of the hollow sphere is concentrated near its surface, while the mass of the solid sphere is distributed uniformly inside it. As a result, the solid sphere has less rotational inertia and, therefore, would roll down the incline with greater acceleration than the hollow sphere.

To maximize the torque produced when you push or pull the knob.

Torque is equal to the applied force times the length of the lever arm (or $\tau = F r$). When there is no lever arm ($r = 0$), the torque ($\tau$) is equal to zero. For example, if you push a door at the edge where the hinges are attached, the door experiences zero torque and no rotation is achieved.

Friction with the floor.

After you throw the bowling ball down the lane, the nonspinning ball experiences a frictional force with the floor, which acts on the ball in the backward direction (see Figure Q8.77). The resulting nonzero torque causes the ball to spin about its center and roll forward.

The first truck on the left.

Start by drawing a vertical arrow through the center of mass of each truck (see Figure Q8.88(b)) to represent the downward force of gravity. The truck tips over when its weight generates a counterclockwise torque about an axis through the left set of wheels.

In Figure Q8.88(b), you see that the weight arrow for the middle truck passes through a point on the ground located between the truck’s left and right wheels. This is also the case for the truck on the right. As a result, these two trucks should be stable. The truck on the left, however, experiences a net torque in the counterclockwise direction, which will cause it to rotate about its left set of wheels and tip over.

Four times more friction.

The centripetal force required to keep an object of mass $m$ moving with a speed $v$ along a circular path of radius $r$ is given by the equation \begin{equation} F = \frac{m v^2}{r} \end{equation} For twice the speed, the new centripetal force required to keep the car on the track is given by \begin{align} F^\prime &= \frac{m\, (2v)^2}{r} \nonumber\\[6pt] &= 4 \left(\frac{m v^2}{r}\right) \nonumber\\[6pt] &= 4 F \nonumber\\[6pt] \end{align} So when the speed of the car doubles, the required force of friction quadruples.

Gravitational attraction exerts a downward force on the coin equal to its weight ($w = mg$). As shown in Figure Q8.94(b), the tabletop exerts an equal upward force, called the normal force ($N = w$), which prevents the coin from falling through the table. The force of friction ($f$) acts parallel to the surface of the table, toward the axis of rotation, and provides the necessary centripetal force to maintain the coin’s rotation and prevent it from sliding off the edge.

The rate of rotation decreases, as given by the law of conservation of angular momentum.

The angular momentum of the rotating system is equal to its rotational inertia ($I$) multiplied by its rotational velocity ($\omega$). As you crawl away from the center of the turntable, you increase your radius of rotation and the rotational inertia of the system (the turntable and you) becomes larger. Since no external torques are involved, the angular momentum of the system remains unchanged: \begin{align} \text{Initial angular momentum} &= \text{final angular momentum} \nonumber\\[6pt] \Rightarrow\quad (I\,\omega)_\text{before} &= (I\,\omega)_\text{after} \end{align} From the equation above, you see that if $I$ increases, $\omega$ must decrease by an equal factor for $I\omega$ to remain constant. Therefore, when you crawl toward the edge of the turntable, you spin more slowly.

It would become shorter.

If the world’s populations moved to the poles, the mass of humans on Earth would become concentrated near Earth’s axis of rotation, which would cause the planet’s overall rotational inertia to decrease slightly. Since Earth’s angular momentum is constant, its rotational velocity would increase slightly (i.e., spin a little faster).

The reading will be low.

The speedometer measures the rotational speed of the wheels and converts it into linear speed for the automobile. Using larger tires, the car will cover a greater distance for every rotation of its wheels. As a result, the speedometer will underestimate the actual speed of the car on the road.

How about $\Sigma \tau = 0$?

A possible situation is shown in Figure Q8.109, where two equal forces act in opposite directions on a disc. In this case, the net force on the disc is equal to zero. However, the disc experiences a nonzero net torque in the clockwise direction. As a result, the disc rotates about its center of mass without changing its location. Mechanical equilibrium is not achieved here because the second condition, $\Sigma \tau = 0$, is not satisfied.

The normal force.

Assume slippery road conditions with almost no friction. The car in Figure Q8.117 is traveling away from you and turning left. In this case, only two forces are acting on the car: the downward force of gravity ($mg$) and the normal force ($N$) perpendicular to the surface of the road.

For the car to safely negotiate the curve, a horizontal centripetal force must act on the car to keep it on track. Note that the vertical force of gravity cannot contribute to the horizontal centripetal force. However, the normal force has a horizontal component ($N_x$) that provides the required centripetal force even in the absence of friction.