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BPO 2024, Теория — Задача 3

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BPO6 Problems 20-24 June 2024. Montenegro · 20 юни 2024 г. · 25 т.

Условие

A conical pendulum: a thread of length l hangs from the suspension point A on a horizontal ceiling line, making angle α with the dashed vertical axis; the ball moves on a dashed horizontal circle whose centre B lies on the axis; point C lies on the axis below B.

A point-like ball of mass $m = 1.0\ \mathrm{g}$ is charged with an amount of positive charge $q = 1.0\ \mathrm{nC}$ . It is suspended by a non-conducting thread of length $l = 0.50\ \mathrm{m}$ . The mass of the thread is negligible. The ball rotates around a vertical axis in a circular path, the thread making an angle $\alpha = 30^{\circ}$ with the vertical axis, as shown in figure.

An additional positive point-like charge $Q = 2.0\ \mathrm{nC}$ can be placed on the vertical axis in three different positions. In the figure those positions are marked with the letters A, B and C. The distance AB is equal with the distance BC.

The friction forces in the system can be neglected. The gravitational acceleration in the system is $g = 9.8\ \mathrm{m/s^2}$ .

A. When the additional charge is at point A:

a) (1 point) Draw the free body diagram for the rotating ball.

b) (2 points) Using the diagram write the equations of motion for the ball.

c) (2 points) Derive the expression and calculate the numerical value for the tension force in the thread.

d) (2 points) Assuming that the circular motion is uniform, derive the formula for the speed of the ball.

e) (2 points) Derive the expression and calculate the period of rotation of the ball. [9 т.]

B. Solve the tasks a, b, c, d and e when the charge $Q$ is at point B. [8 т.]

C. Solve the tasks a, b, c, d and e when the charge $Q$ is at point C. [8 т.]

Отговори

Решение

A ball of mass $m = 1\ \mathrm{g}$ is charged with an amount of positive charge $q = 1\ \mathrm{nC}$ . It is suspended by a non – conducting thread of length $l = 0.5\ \mathrm{m}$ . The mass of the thread is negligible. The thread makes an angle $\alpha = 30^{\circ}$ with the vertical axis, as shown in figure.

An additional positive charge $Q = 2\ \mathrm{nC}$ can be placed on the vertical axis in three different positions. In the figure those positions are marked with letters A, B and C. The distance AB is equal to BC. The ball rotates around a vertical axis in a circular path. The friction forces in the system can be neglected. The gravitational acceleration in the system is $g = 9.81\ \mathrm{m/s^2}$ . Constant $k = 9 \cdot 10^{9}\ \mathrm{Nm^2/C^2}$ .

A. When the additional charge is at point A:

1. Draw the free body diagram for the rotating ball. (1 point)

Free body diagram with the charge Q at point A (red dot at the suspension point on the ceiling line): thread of length l at angle α to the dashed vertical axis; forces on the ball: tension F_t along the thread upward, Coulomb force F_c along the thread away from A, weight F_g = mg downward, dashed red centrifugal force F_cf = ma_n outward; the acceleration a_n points toward the axis along the horizontal circle.

Note:

Since centrifugal force is a fictitious force acting on a ball, it can be omitted from the free body diagram. In that case the marking scheme is the same, i.e. the maximal number of points that can be given without centrifugal force in the diagram is the same.

2. Using the diagram write the equations of motion of the ball. (2 points)

The same free body diagram for the charge Q at point A: thread l at angle α, forces F_t, F_c, F_g = mg, dashed red F_cf = ma_n, and acceleration a_n toward the axis.

The charge $q$ is at rest in its own reference frame. Therefore (0.5 points):

\vec{F}_{\mathrm{g}} + \vec{F}_{\mathrm{cf}} + \vec{F}_{\mathrm{c}} + \vec{F}_{\mathrm{t}} = \vec{0}

In this case the centrifugal force must be included. By decomposing the vectors along horizontal and perpendicular axes (0.5 points), we get the equations of motion of the ball: (1 point)

ma_n + k\frac{qQ}{l^2}\sin\alpha = F_{\mathrm{t}}\sin\alpha

mg + k\frac{qQ}{l^2}\cos\alpha = F_{\mathrm{t}}\cos\alpha

3. Determine the tension force in the thread. (2 points)

From the equation above, it follows that

F_{\mathrm{t}} = \frac{mg}{\cos\alpha} + k\frac{qQ}{l^2} = 11.3\ \mathrm{mN}

4. Assuming that the circular motion is uniform, derive the formula for the linear speed of the ball. (2 points)

Using the equation (1 point)

\frac{mv^2}{l\sin\alpha} + k\frac{qQ}{l^2}\sin\alpha = F_{\mathrm{t}}\sin\alpha

it can be obtained (1 point)

v = \sqrt{\frac{gl\sin^2\alpha}{\cos\alpha}}

5. Determine the period of rotation of the ball. (2 points)

The distance travelled by the ball during one period is equal to the circumference of the circle $s = 2\pi l\sin\alpha$ . (1 point)

Therefore, the period of rotation is (1 point)

T = \frac{s}{v} = 2\pi\sqrt{\frac{l\cos\alpha}{g}} = 1.322\ s

B. Solve the problems 1 – 5 when the charge $Q$ is at point B.

1. Draw the free body diagram for the rotating ball. (1 point)

Free body diagram with the charge Q at point B (red dot at the centre of the circle): thread l at angle α, tension F_t along the thread, horizontal Coulomb force F_c outward, dashed red F_cf = ma_n outward, weight F_g = mg downward, acceleration a_n toward the centre.

Note:

2. Using the diagram write the equations of motion of the ball. (2 points)

The charge $q$ is at rest in its own reference frame. Therefore (0.5 points):

\vec{F}_{\mathrm{g}} + \vec{F}_{\mathrm{cf}} + \vec{F}_{\mathrm{c}} + \vec{F}_{\mathrm{t}} = \vec{0}

In this case the centrifugal force must be included. By decomposing the vectors along horizontal and perpendicular axes (0.5 points), we get the equations of motion of the ball: (1 point)

ma_n + k\frac{qQ}{l^2\sin^2\alpha} = F_{\mathrm{t}}\sin\alpha

mg = F_{\mathrm{t}}\cos\alpha

3. Determine the tension force in the thread. (1 point)

From the equation above, it follows that

F_{\mathrm{t}} = \frac{mg}{\cos\alpha} = 11.32\ \mathrm{mN}

4. Assuming that the circular motion is uniform, derive the formula for the linear speed of the ball. (2 points)

Using the equation (1 point)

\frac{mv^2}{l\sin\alpha} + k\frac{qQ}{l^2\sin^2\alpha} = F_{\mathrm{t}}\sin\alpha

it can be obtained (1 point)

v = \sqrt{\left(mg\mathrm{tg}\alpha - k\frac{qQ}{l^2\sin^2\alpha}\right)\frac{l\sin\alpha}{m}}

5. Determine the period of rotation of the ball. (2 points)

The distance travelled by the ball during one period is equal to the circumference of the circle $s = 2\pi l\sin\alpha$ . (1 point)

Therefore, the period of rotation is (1 point)

T = \frac{s}{v} = 2\pi\sqrt{\frac{ml\sin\alpha}{\left(mg\mathrm{tg}\alpha - k\frac{qQ}{l^2\sin^2\alpha}\right)}} = 1.3205\ s

C. Solve the problems 1 - 5 when the charge $Q$ is at point C.

1. Draw the free body diagram for the rotating ball. (1 point)

Free body diagram with the charge Q at point C (red dot on the axis below the circle): thread l at angle α, tension F_t along the thread, Coulomb force F_c directed up and outward away from C, dashed red F_cf = ma_n outward, weight F_g = mg downward, acceleration a_n toward the axis.

Note:

2. Using the diagram write the equations of motion of the ball. (2 points)

The charge $q$ is at rest in its own reference frame. Therefore (0.5 points):

\vec{F}_{\mathrm{g}} + \vec{F}_{\mathrm{cf}} + \vec{F}_{\mathrm{c}} + \vec{F}_{\mathrm{t}} = \vec{0}

In this case the centrifugal force must be included. By decomposing the vectors along horizontal and perpendicular axes (0.5 points), we get the equations of motion of the ball: (1 point)

ma_n + k\frac{qQ}{l^2}\sin\alpha = F_{\mathrm{t}}\sin\alpha

mg = \left(F_{\mathrm{t}} + k\frac{qQ}{l^2}\right)\cos\alpha

3. Determine the tension force in the thread. (1 point)

From the equation above, it follows that

F_{\mathrm{t}} = \frac{mg}{\cos\alpha} - k\frac{qQ}{l^2} = 11.328\ \mathrm{mN}

4. Assuming that the circular motion is uniform, derive the formula for the linear speed of the ball. (2 points)

Using the equation (1 point)

\frac{mv^2}{l\sin\alpha} + k\frac{qQ}{l^2}\sin\alpha = F_{\mathrm{t}}\sin\alpha

it can be obtained (1 point)

v = \sqrt{\left(F_{\mathrm{t}} - k\frac{qQ}{l^2}\right)\frac{l}{m}}\ \sin\alpha

5. Determine the period of rotation of the ball. (2 points)

The distance travelled by the ball during one period is equal to the circumference of the circle $s = 2\pi l\sin\alpha$ . (1 point)

Therefore, the period of rotation is (1 point)

T = \frac{s}{v} = 2\pi\sqrt{\frac{ml}{\left(F_{\mathrm{t}} - k\frac{qQ}{l^2}\right)}} = 1.3201\ s

Оригинал в Архива: problems.pdf · официални решения: solutions.pdf