Review problem. A 1.00 -g cork ball with charge 2.00μC is suspended vertically on a 0.500 -m-lon

Review problem. A 1.00 -g cork ball with charge 2.00μC is...

Review problem. A 1.00 -g cork ball with charge 2.00$\mu \mathrm{C}$ is suspended vertically on a 0.500 -m-long light string in the presence of a uniform, downward-directed electric field of magnitude $E=1.00 \times 10^{5} \mathrm{N} / \mathrm{C}$ . If the ball is displaced slightly from the vertical, it oscillates like a simple pendulum. (a) Determine the period of this oscillation. (b) Should gravity be included in the calculation for part (a)? Explain.

Review problem. A 1.00 -g cork ball with charge 2.00$\mu \mathrm{C}$ is suspended vertically on a 0.500 -m-long light string in the presence of a uniform, downward-directed electric field of magnitude $E=1.00 \times 10^{5} \mathrm{N} / \mathrm{C}$ . If the ball is displaced slightly from the vertical, it oscillates like a simple pendulum. (a) Determine the period of this oscillation.
(b) Should gravity be included in the calculation for part
(a)? Explain.

Key Concepts

Simple Pendulum Oscillation

A simple pendulum consists of a mass suspended by a string where the restoring force for small displacements is proportional to the sine of the angle. The motion is periodic, and for small angles the period can be approximated by the formula T = 2??(L/g), with L being the length of the pendulum and g the acceleration due to gravity. This concept underlies many oscillatory systems, connecting mechanics with simple harmonic motion.

Small Angle Approximation

When oscillations involve small angular displacements, the sine of the angle can be approximated by the angle in radians. This simplification linearizes the equations of motion for a pendulum, leading to a direct formula for the period. It is a central concept in analyzing oscillatory systems because it allows complex, nonlinear behavior to be studied using simpler linear equations.

Electric Force on Charges in a Uniform Field

A charged object in a uniform electric field experiences a force equal to the product of its charge and the electric field strength. This force can modify the equilibrium position of the object when it is suspended, as it adds to or subtracts from the gravitational force. However, for small oscillations about the new equilibrium position, this constant force does not change the nature of the restoring force that governs the oscillation period.

Effective Acceleration in Oscillatory Motion

When additional constant forces are present, such as those due to an electric field, they shift the equilibrium position but do not generally alter the restoring force for small oscillations. The system oscillates about a new equilibrium with an effective acceleration that is still related to gravity. This concept explains why gravity remains a key factor in calculating the period of the pendulum's oscillation.

Transcript

00:01 So in this problem we're dealing with the influence of electrostatic force caused by an electrostatic field on the physical bodies and the relationship between the electrostatic force and forces coming from other sources like gravitational source in this case.

00:21 So here we have a pendulum that is being influenced by an electrostatic field.

00:26 So it is...

00:28 Pendulom will oscillate normally, for example, like this.

00:34 But the angle of teeth is very small, and so this will be a simpler form of pendulam oscillation.

00:43 And what we need to do first is to find the net force which is influencing this.

00:51 This mandolome, in this case, both forces are downwards, so the component in this case will be basically f net will be equal to m g plus and in this case we can say that the since this is unit of the unit to charge on the these two microcolums and this charge basically will be we need to include the force that is due to electrostatic field and this will be this means multiplying, multiplying the charge by the electrostatic field.

01:53 And then the component, which is sign of minus theta.

01:58 Why minus theta? well, because, as you can see, the angle is the original theta is between this and this line, and this theta is now upwards.

02:11 So we need to include the minus sine.

02:12 And since sine is not a symmetric function, if we said the minus theta, this means the whole expression becomes minus.

02:24 And this is minus mg plus q times e sine is of theta.

02:38 And also, we don't really need to keep this vector notation because we only need the magnitude of this vector or the net force.

02:49 Now when we substitute the values to obtain, this is actually 0 .0 .0.

02:55 0981 because this is 1 gram which is 10 to negative third power kilograms plus 0 .2 because the charge was in given in microcolumns and the electric field is on 10 to 5th power of neutrons per column i think yes so and the sign function can be approximated with theta why because sign of theta is approximated just theta for very small angles and this works here as well therefore now we need to express the net force in another way so every force is basically mass times acceleration this case acceleration is related to the angular acceleration which is m times alpha times the radius are the radius of this radius will be basically the length of the string this is r here and this equals to minus m g plus kub theta so there is a relationship between for this small angles there we know that second derivative of theta over t is basically minus omega squared but also this this kind of this is from the standard differential equation for the in billum for the small angle of displacement.

04:36 But also this is equal to alpha...

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