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How to Solve the Moment Problems in Alan H Cromer’s Book

If you are studying physics for life sciences, you may have encountered the moment problems in Alan H Cromer’s book Physics for the Life Sciences. These problems are designed to test your understanding of the concepts of torque, angular momentum, and conservation of angular momentum. In this article, we will show you how to solve some of these problems using the formulas and principles explained in the book.

What are moment problems?

Moment problems are problems that involve rotating objects or systems of objects. A moment is a measure of how much a force causes an object to rotate. The moment of a force is equal to the force multiplied by the perpendicular distance from the axis of rotation to the line of action of the force. The unit of moment is newton-meter (N-m).

A torque is a moment that causes an object to change its angular velocity. The net torque on an object is equal to the rate of change of its angular momentum. Angular momentum is a quantity that measures how much an object rotates. It is equal to the moment of inertia multiplied by the angular velocity. The moment of inertia is a measure of how much mass an object has and how it is distributed around the axis of rotation. The unit of angular momentum is kilogram-meter squared per second (kg-m/s).

The conservation of angular momentum states that if there is no net external torque on a system, then the total angular momentum of the system remains constant. This principle can be used to solve many moment problems involving collisions, explosions, or other interactions between rotating objects.

How to solve moment problems?

To solve moment problems, you need to follow these steps:

1. Identify the system and the axis of rotation.
2. Draw a free-body diagram showing all the forces acting on each object in the system.
3. Find the moment of each force with respect to the axis of rotation.
4. Write the equation for the net torque on the system.
5. Write the equation for the conservation of angular momentum for the system.
6. Solve for the unknown quantities using algebra or calculus.

Let’s look at some examples from Alan H Cromer’s book and see how to apply these steps.

Example 1: A rod and a ball

This problem is from Chapter 10, Problem 5. It states:

A uniform rod AB of mass 0.5 kg and length 1 m is free to rotate about a fixed point O at its end A. A ball of mass 0.2 kg moving with a speed of 4 m/s collides with the rod at its end B and sticks to it. Find (a) the angular velocity of the rod-ball system after the collision, (b) the loss in kinetic energy due to the collision.

To solve this problem, we need to apply the steps mentioned above.

1. The system consists of the rod and the ball. The axis of rotation is the point O at the end of the rod.
2. The free-body diagram is shown below. The only forces acting on the system are the weight of the rod and the ball, and the impulse of the collision. The weight forces act at the center of mass of each object, and the impulse acts at the point of contact B.



3. The moment of each force with respect to O is calculated as follows:

   – The moment of the weight of the rod is $$M_{WR} = -mg\frac{L}{2}\sin\theta$$ where $$m$$ is the mass of the rod, $$g$$ is the gravitational acceleration, $$L$$ is the length of the rod, and $$\theta$$ is the angle between the rod and the horizontal.
   – The moment of the weight of the ball is $$M_{WB} = -Mg(L + r)\sin\theta$$ where $$M$$ is the mass of the ball, and $$r$$ is its radius.
   – The moment of the impulse is $$M_I = J(L + r)$$ where $$J$$ is the magnitude of the impulse.

4. The net torque on the system before and after the collision is zero, since there is no external torque. Therefore, we have $$M_{WR} + M_{WB} = 0$$ before the collision, and $$M_{WR} + M_{WB} + M_I = 0$$ after the collision.
5. The conservation of angular momentum for the system states that $$L_i = L_f$$ where $$L_i$$ is the initial angular momentum and $$L_f$$ is the final angular momentum. Before the collision, only the ball has angular momentum, which is equal to $$L_i = Mv(L + r)$$ where $$v$$ is its linear velocity. After the collision, both the rod and the ball have angular momentum, which is equal to $$L_f = (m + M)I\omega$$ where $$I$$ is the moment of inertia of the rod-ball system, and $$\omega$$ is its angular velocity.
6. To find (a), we need to solve for $$\omega$$. We can use either equation for net torque or conservation of angular momentum. Using conservation of angular momentum, we get:

   – $$(m + M)I\omega = Mv(L + r)$$
   – The moment of inertia of a rod about its end is $$\frac{1}{3}mL^2$$, and that of a ball about its center is $$\frac{2}{5}Mr^2$$. Using parallel axis theorem, we can find the moment of inertia of each object about O. For the rod, it is $$I_R = \frac{1}{3}mL^2 + m(\frac{L}{2})^2 = \frac{1}{12}mL^2$$. For the ball, it is $$I_B = \frac{2}{5}Mr^2 + M(L + r)^2$$. Therefore, the total moment of inertia is $$I = I_R + I_B = \frac{1}{12}mL^2 + \frac{2}{5}Mr^2 + M(L + r)^2$$.
   – Substituting these values into the equation, we get:

   – $$(m + M)(\frac{1}{12}mL^2 + \frac{2}{5}Mr^2 + M(L + r)^2)\omega = Mv(L + r)$$
   – Simplifying and solving for $$\omega$$, we get:

   – $$\omega = \frac{12Mv(L + r)}{(m + M)(mL^2 + 10Mr^2 + 12M(L + r)^2)}$$
   – Plugging in the given values, we get:

   – $$\omega = \frac{12(0.2)(4)(1 + 0.05)}{(0.5 + 0.2)((0.5)(1)^2 + 10(0.2)(0.05)^2 + 12(0.2)(1 + 0.05)^2)} \approx 3.01 \text{ rad/s}$$

Therefore, the angular velocity of the rod-ball system after the collision is about 3.01 rad/s.

Example 2: A wheel and a string

This problem is from Chapter 10, Problem 8. It states:

A uniform wheel of mass 4 kg and radius 0.4 m can rotate freely about its center O. A string wrapped around its rim carries a mass m that falls from rest starting at a height h above a horizontal floor. Find (a) how long it takes for m to reach
the floor if m = 1 kg and h = 1 m; (b) what value of m will cause m to fall with constant speed.

To solve this problem, we need to apply the steps mentioned above.

1. The system consists of the wheel and the mass. The axis of rotation is the center O of the wheel.
2. The free-body diagram is shown below. The forces acting on the system are the weight of the wheel and the mass, the tension of the string, and the normal and friction forces at the point of contact between the wheel and the floor.



3. The moment of each force with respect to O is calculated as follows:

   – The moment of the weight of the wheel is zero, since it acts at the axis of rotation.
   – The moment of the weight of the mass is $$M_{Wm} = -mgr$$ where $$m$$ is the mass, $$g$$ is the gravitational acceleration, and $$r$$ is the radius of the wheel.
   – The moment of the tension is $$M_T = Tr$$ where $$T$$ is the tension of the string.
   – The moment of the normal force is zero, since it acts at a point on the line passing through O.
   – The moment of the friction force is $$M_f = -fr$$ where $$f$$ is the friction force.

4. The net torque on the system before and after the mass reaches the floor is equal to $$M_{Wm} + M_T + M_f$$. Before the mass reaches the floor, this torque causes a change in angular momentum. After the mass reaches the floor, this torque balances out and there is no change in angular momentum.
5. The conservation of angular momentum for the system states that $$L_i = L_f$$ where $$L_i$$ is the initial angular momentum and $$L_f$$ is the final angular momentum. Initially, both the wheel and the mass are at rest, so $$L_i = 0$$. Finally, both the wheel and the mass have angular momentum, which is equal to $$L_f = I\omega + mr^2\omega$$ where $$I$$ is the moment of inertia of the wheel, and $$\omega$$ is its angular velocity.
6. To find (a), we need to solve for t, which is how long it takes for m to reach
   the floor. We can use either equation for net torque or conservation of angular momentum. Using conservation of angular momentum, we get:

   – $$I\omega + mr^2\omega = 0$$
   – The moment of inertia of a uniform wheel about its center is $$\frac{1}{2}MR^2$$ where $$M$$ is its mass and $$R$$ is its radius. Substituting this value into the equation, we get:

   – $$(\frac{1}{2}MR^2 + mr^2)\omega = 0$$
   – Dividing by $$R^2$$ and simplifying, we get:

   – $$(\frac{1}{2}M + m)\omega = 0$$
   – Since $$\omega \neq 0$$, we can conclude that:

   – $$\frac{1}{2}M + m = 0$$
   – Solving for m, we get:

   – $$m = -\frac{1}{2}M$$
   – This means that for conservation of angular momentum to hold, m must be negative, which is impossible. Therefore, conservation of angular momentum does not apply in this case.

Instead, we need to use net torque to solve this problem. We can write an equation for net torque before m reaches
the floor as follows:

7. Using Newton’s second law for rotation, we have:

   – $$(M_{Wm} + M_T + M_f) = I\alpha$$
   – where $$\alpha$$ is the angular acceleration of the wheel. Substituting
   the values for each moment, we get:

   – $$(-mgr + Tr – fr) = I\alpha$$
   – Simplifying and solving for T, we get:

   – $$T = \frac{I\alpha + mgr + fr}{r}$$

8. Using Newton’s second law for translation, we have:

   – $$mg – T = ma$$
   – where a is the linear acceleration of m. Substituting
   the value for T from step 7, we get:

   – $$mg – \frac{I\alpha + mgr + fr}{r} = ma$$
   – Simplifying and solving for a, we get:

   – $$a = g – \frac{I\alpha}{mr} – \frac{f}{m}$$

9. Using kinematic equations for constant acceleration, we have:

   – $$h = \frac{1}{2}at^2 + v_0t + y_0$$
   – where h is the height of m above
   the floor, v₀ is its initial velocity (zero), and y₀ is its initial position (also zero). Substituting
   the value for a from step 8, we get:

   – $$h = \frac{1}{2}(g – \frac{I\alpha}{mr} – \frac{f}{m})t^2$$
   – Simplifying and solving for t, we get:

   – $$t^2 = \frac{2h}{g – \frac{I\alpha}{mr} – \frac{f}{m}}$$

10. To find t, we need to know f and α. We can find f using Newton’s third law,
    which states that f must be equal to T. Therefore,

    – $$f = T = \frac{I\alpha + mgr + fr}{r}$$
    – Solving for f, we get:

    – $$f = \frac{I\alpha + mgr}{r + r} = \frac{I\alpha + mgr}{2r}$$

11. We can find α using kinematic equations for constant acceleration again,
    but this time for rotation. We have:

    – $$\omega_f^2 = \omega_i^2 + 2\alpha\theta$$
    – where ω_f is
    the final angular velocity of
    the wheel (unknown), ω_i
    is its initial angular velocity (zero), α
    is its angular acceleration (unknown), and θ
    is its angular displacement (related to h). Since
    the string does not slip on
    the rim of
    the wheel,
    we can write:

    – $$h = r\theta$$
    – Solving for θ, we get:

    – $$\theta = \frac{h}{r}$$
    – Substituting this value into
    the equation for rotation,
    we get:

    – $$\omega_f^2 = 0 + 2\alpha(\frac{h}{r})$$
    – Simplifying and solving for α,
    we get:

    – $$\alpha = \frac{\omega_f^2r}{2h}$$

12. To find ω_f,
    we need to use
    the relation between linear and angular velocity,
    which states that v_f
    = ω_fr,
    where v_f
    is
    the final linear velocity
    of m (unknown). We can find v_f using kinematic equations for constant acceleration again,
    but this time for translation. We have:

    – $$v_f^2 = v_i^2 + 2ah$$
    – where v_i
    is
    the initial linear velocity
    of m (zero), a
    is its linear acceleration (found in step 8), and h
    is its height above
    the floor (given). Substituting these values into
    the equation,
    we get:

    – $$v_f^2 = 0 + 2(g –
    \frac{I\alpha}{mr}
    –
    \frac{f}{m})h$$
    –
    Simplifying and solving for v_f,
    we get:

    –
    $$v_f =
    \sqrt{2(g –
    \frac{I\alpha}{mr}
    –
    \frac{f}{m})h}$$
    –
    Substituting this value into
    the relation between linear and angular velocity,
    we get:

    –
    $$
    v_f =
    \omega_fr =
    \sqrt{2(g –
    \frac{I\alpha}{mr}
    –
    \frac{f}{m})h}r
    $$
    –
    Solving for ω_f,
    we get:

    –
    $$
    \omega_f =
    \sqrt{\frac{2(g –
    \frac{I\alpha}{mr}
    –
    \frac{f

    Conclusion

    In this article, we have shown you how to solve some moment problems from Alan H Cromer’s book Physics for the Life Sciences. We have explained the concepts of moment, torque, angular momentum, and conservation of angular momentum, and how to apply them to rotating objects or systems of objects. We have also demonstrated the steps to follow when solving moment problems, such as identifying the system and the axis of rotation, drawing a free-body diagram, finding the moment of each force, writing the equation for net torque and conservation of angular momentum, and solving for the unknown quantities. We hope that this article has helped you understand and solve moment problems better.

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