Circular Motion and Gravitation (Chapter # 07) Physics 10th - Question Answers

 Physics 10th - Question Answers

Circular Motion and Gravitation (Chapter # 07)

Q.1: Define Circular Motion and Give Any Two Examples.

Ans: Circular Motion: When a body moves along the circumference of a circle, it is called circular motion.

Examples:

• The rotation of Earth and other planets around the Sun.
• The wheel of a moving bicycle.

Q.2: Define Uniform Circular Motion. Explain It with One Example.

Ans: Uniform Circular Motion: When a body moves in a circular path at a constant speed, it is said to be in uniform circular motion.

Explanation:

• At any point on the circle, the direction of velocity is along the tangent to the circle at that point.
• If a body is freed from force at any point, it will move tangentially to the circle.
• As the body moves along the circumference of the circle, there must be a force directed towards the center of the circle, known as the centripetal force.

Example: Consider a body revolving in a circle of radius $r$. Suppose the body is initially at point $P$. After some time, it moves to a new position. The angle $\angle POP$ or $\theta$ subtended at the center of the circle represents the turning of the body during angular displacement.

Unit: The unit of angular displacement is the radian.

Definition of Radian: One radian is the angle subtended at the center of a circle by an arc whose length is equal to the radius of the circle.

$$1 \text{ radian} = 57.30^\circ$$

Q.3: Write Down the Names of Two Forces Acting on a Body Moving in a Circular Path.

Ans: There are two forces acting on a body when it moves in a circular path:

• Centripetal Force: $$F_c = \frac{mv^2}{r}$$
• Centrifugal Force: $$F_c = -\frac{mv^2}{r}$$

Q.4: Define Centripetal Acceleration. Derive the Equation $a_c = \frac{v^2}{r}$.

Ans: Centripetal Acceleration: The change of velocity along the circumference of a circle is called centripetal acceleration. It is denoted by $a_c$. The S.I. unit of centripetal acceleration is $\text{m/s}^2$.

Derivation of Equation $a_c = \frac{v^2}{r}$: Consider a body of mass $m$ moving with a constant speed $v$ around a circle of radius $r$.

• At any instant, the direction of velocity is along the tangent to the circular path.
• As the body moves along the circumference, the direction of velocity continuously changes, indicating the presence of acceleration.

The magnitude of this acceleration is directly proportional to the square of the velocity:

$$a_c \propto v^2 \quad \text{(i)}$$

The magnitude of acceleration is inversely proportional to the radius of the circle:

$$a_c \propto \frac{1}{r} \quad \text{(ii)}$$

On combining equations (i) and (ii), we get:

$$a_c \propto \frac{v^2}{r}$$

Thus:

$$a_c = \frac{v^2}{r}$$

Q.5: Define Centripetal Force. Give Any Two Examples.

Ans: Centripetal Force: The force that acts towards the center along the radius of a circular path on which the body is moving with uniform velocity is called centripetal force.

Examples:

• When a bucket filled with water is rotated in a circle, the water does not fall out, even when the bucket is at its highest point. The centripetal force is provided by the weight of the water acting vertically downward, preventing the water from falling.
• In the case of the moon orbiting the Earth, the centripetal force is provided by the gravitational force of attraction between the Earth and the moon. Similarly, the force of attraction from the Sun on the planets provides the necessary centripetal force, making the planets revolve around the Sun.

Q.6: Derive the Equation $F_c = \frac{mv^2}{r}$.

Ans: From Newton’s second law of motion, we have:

$$F_c = m a_c \quad \text{(i)}$$

Where $a_c$ is the centripetal acceleration produced by the centripetal force $F_c$. We know:

$$a_c = \frac{v^2}{r}$$

Substituting this into equation (i), we get:

$$F_c = m \left(\frac{v^2}{r}\right)$$

Another Method: It has been found experimentally that:

• Centripetal force is directly proportional to the mass of the object moving in a circular path: $$F_c \propto m \quad \text{(i)}$$
• Centripetal force is directly proportional to the square of the velocity of the object: $$F_c \propto v^2 \quad \text{(ii)}$$
• Centripetal force is inversely proportional to the radius of the circle: $$F_c \propto \frac{1}{r} \quad \text{(iii)}$$

On combining equations (i), (ii), and (iii), we have:

$$F_c \propto \frac{mv^2}{r}$$

Thus:

$$F_c = \frac{mv^2}{r}$$

Q.7: Write Down the Factors on Which Centripetal Force Depends.

Ans: The following factors influence centripetal force:

• Centripetal force increases with the increasing mass of the body.
• It increases with the square of the velocity of the body.
• It decreases as the radius of the circular path increases.

Q.8: Write Down the Factors on Which Centripetal Acceleration Depends.

Ans: Centripetal acceleration depends on the following factors:

• It is directly proportional to the square of the velocity.
• It is inversely proportional to the radius of the circle.

Q.9: Define Centrifugal Force. Give Two Examples.

Ans: Centrifugal force is the force directed away from the center of a circle. It is the opposite and equal reaction to centripetal force.

$$F_c = -\frac{mv^2}{r}$$

Examples:

• When a stone tied to a string is whirled in a circle, the force keeping the stone in circular motion pulls it inward along the string.
• According to Newton’s third law of motion, the stone exerts an equal force outward on the hand. This outward force is called centrifugal force.

Q.10: What Is Banking of a Road? What Are Its Advantages?

Ans: The outer edges of curved roads are made higher than the inner edges. This is called banking of the road.

Advantages of Banking a Road:

• When a vehicle exceeds a certain speed limit while turning a corner, the centripetal force may not be sufficient to keep the vehicle in circular motion. The vehicle may tend to slide outward from the center.
• A banked road helps keep the vehicle safely on the curve. Speed limits are posted around curves for safe driving.

Q.11: Define Centrifuge. Give Some of Its Applications.

Ans: Centrifuge: A centrifuge is a device that separates heavier particles from lighter particles in a liquid. It is a cylindrical vessel in which a liquid is rotated at high speed by an electric motor.

Applications of a Centrifuge:

• Wet clothes are spun at high speed in washing machine dryers. The walls of the dryers have numerous holes. When the dryer runs, water is separated from the clothes by centrifugal force and exits through the holes, drying the clothes.
• Sugar crystals are separated from molasses using centrifuges.
• A cream separator also acts as a centrifuge. The cream collects near the axis, while skimmed milk moves away from the axis of rotation.

Q.12: State Newton’s Law of Gravitation. Derive the Equation $F = G\frac{m_1 m_2}{r^2}$.

Ans: Statement: Every body in the universe attracts every other body with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

Derivation of Equation: If $F$ is the force between two bodies of masses $m_1$ and $m_2$, and $r$ is the distance between their centers:

• According to the law of gravitation, the force is directly proportional to the product of their masses:

  $$F \propto m_1 m_2 \quad \text{(i)}$$

• The force is inversely proportional to the square of the distance between them:

  $$F \propto \frac{1}{r^2} \quad \text{(ii)}$$

On combining equations (i) and (ii), we get:

$$F \propto \frac{m_1 m_2}{r^2}$$

Thus:

$$F = G\frac{m_1 m_2}{r^2}$$

Where $G$ is the gravitational constant, with a value of $6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2$.

Q.13: Derive an Expression for the Determination of the Value of Gravitational Constant (G) by Cavendish Method.

Ans: The value of $G$ was determined experimentally by Henry Cavendish in 1798 using an instrument called the Cavendish balance. The Cavendish balance consists of two small spheres of mass $m$, mounted at opposite ends of a light horizontal rod of length $d$, suspended at its center by a fine vertical fiber.

When two large spheres of mass $M$ are brought near the small spheres at a distance $r$, the force of gravitational attraction between the spheres produces torque, twisting the fiber. The twist in the fiber is proportional to the magnitude of the torque.

If $G\frac{mM}{r^2}$ is the gravitational force between the spheres and $F \times d$ is the torque producing the twist, then:

$$G\frac{mM}{r^2} \times d = C \theta$$

Where "C" is a proportionality constant, depending on the material of the fiber. The value of $C$ can be determined experimentally.

By substituting the known quantities, the value of $G$ can be calculated:

$$G = \frac{r^2 C \theta}{M m d}$$

The value of $G$ calculated by the Cavendish method is $6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2$.

Q.14: Using Newton’s Law of Gravitation, Find the Mass of the Earth.

Ans: Consider a body of mass $m$ placed on the surface of the Earth. Let $M_e$ be the mass of the Earth and $R_e$ its radius. According to Newton’s Law of Gravitation, the force with which the Earth attracts the body towards its center is given by:

$$F = G \frac{m M_e}{R_e^2} \quad \text{(i)}$$

But the force of attraction is equal to the weight of the body:

$$F = W = m g \quad \text{(ii)}$$

By comparing equations (i) and (ii):

$$G \frac{m M_e}{R_e^2} = m g$$

Solving for $M_e$, the mass of the Earth can be determined.

Q.15: Explain How the Value of "g" Decreases with a Change in Altitude.

Ans: We consider a body of mass $m$ on the Earth's surface. The gravitational force of attraction between the body and the Earth is equal to the weight of the body:

$$F = W = \frac{G M_e m}{R_e^2}$$

Thus:

$$g = \frac{G M_e}{R_e^2} \quad \text{(i)}$$

If the body is moved up to a height $h$ above the Earth's surface, the acceleration due to gravity at a distance $R_e + h$ from the center of the Earth, denoted as $g_h$, is given by:

$$g_h = \frac{G M_e}{(R_e + h)^2} \quad \text{(ii)}$$

Dividing equation (ii) by equation (i), we get:

$$\frac{g_h}{g} = \frac{\frac{G M_e}{(R_e + h)^2}}{\frac{G M_e}{R_e^2}}$$

This simplifies to:

$$\frac{g_h}{g} = \frac{R_e^2}{(R_e + h)^2}$$

The derived equation shows that as the value of $h$ increases, the value of gravitational acceleration $g$ decreases. This indicates that the acceleration due to gravity decreases with altitude.

Q.16: What Is a Satellite? What Are Natural and Artificial Satellites?

Ans: Satellite: An object revolving around a planet in a fixed orbit is called a satellite. There are two types of satellites:

• Natural Satellites: These exist naturally and revolve around planets. An example is the Moon, which revolves around the Earth.

• Artificial Satellites: These are man-made satellites sent by scientists to orbit the Earth. They serve various purposes, such as telecommunication, space research, and weather monitoring.

Q.17: Derive an Expression for the Orbital Velocity of an Artificial Satellite.

Ans: Orbital Velocity: Let a satellite of mass $m$ move in an orbit of radius $r$ with velocity $v$. The gravitational force of attraction between the satellite and the Earth provides the necessary centripetal force.

Gravitational Force:

$$\frac{G M_e m}{r^2}$$

Centripetal Force:

$$\frac{mv^2}{r}$$

Since the gravitational force is equal to the centripetal force:

$$\frac{G M_e m}{r^2} = \frac{mv^2}{r}$$

Rearranging, we get:

$$v^2 = \frac{G M_e}{r}$$

Taking the square root:

$$v = \sqrt{\frac{G M_e}{r}}$$

If the satellite is close to the Earth's surface ($r = R_e$):

$$v = \sqrt{g R_e}$$

Given $g = 9.8 \, \text{m/s}^2$ and $R_e = 6.4 \times 10^6 \, \text{m}$:

$$v = \sqrt{9.8 \times 6.4 \times 10^6}$$

Q.18: Is It Possible for a Body to Be Accelerated If Its Speed Is Constant?

Ans: Yes, it is possible. If a body moves with a constant speed but changes its direction, it has acceleration. A body moving at a uniform speed in a circular path experiences acceleration because the centripetal force acts perpendicular to the instantaneous velocity, changing the direction of the velocity without altering its magnitude.

Q.19: Why Is the Acceleration of a Body Moving Uniformly in a Circle Directed Towards the Center?

Ans: According to Newton's Second Law of Motion, acceleration is produced in the direction of the applied force. For a body moving uniformly in a circle, the force (centripetal force) is always directed towards the center of the circle, causing the acceleration to also be directed towards the center.

Q.20: When a Car Turns Left, in Which Direction Do the Occupants Tend to Fall?

Ans: A force is required to keep an object moving in a circle, directed towards the center (centripetal force). When a car turns left, the friction between the road and the tires provides this force. The occupants tend to fall to the right due to inertia. If the road is slippery or the tires are worn out, there may be insufficient centripetal force, causing the car to skid left. Banking of the road (with the outer edge higher than the inner edge) helps the car turn safely at prescribed speed limits.

Q.21: How Would the Values of $g$ and $G$ Be Affected If the Mass of the Earth Becomes Four Times Greater?

Ans: The value of $g$ (gravitational acceleration) is directly proportional to the mass of the Earth. If the mass of the Earth increases, $g$ would also increase proportionally. If the mass decreases, $g$ would decrease as well. The gravitational constant $G$

$G$ remains unchanged as it is a universal constant.

If the mass of the Earth becomes four times greater, the value of $g$ will also increase four times. However, the gravitational constant $G$ remains the same throughout the universe. Its value is:

$$G = 6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2$$

Q.22: Write Down the Difference Between "G" and "g".

G	g
Represents the gravitational constant.	Represents acceleration due to gravity.
It is a scalar quantity.	It is a vector quantity.
It remains constant.	It varies from place to place.
Value: $6.67 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2$.	Value: $9.8 \, \text{m/s}^2$.

Q.23: Write Down the Difference Between Centripetal Force and Centrifugal Force.

Centripetal Force	Centrifugal Force
It is directed towards the center.	It is directed away from the center.
Acts on the moving object.	Acts on the object providing the centripetal force.
Tends to keep the body moving in a circle.	Tends to keep the body moving in a straight line.

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