Uniform Circular Motion Questions in English

(A) Diameter of circular track $(D) = 200\, m$.
Radius of circular track $(r) = D / 2 = 100\, m$.
Time taken for one round $(t) = 40\, s$.
Total time given $= 2\, min\, 20\, s = 140\, s$.
Number of rounds $= 140 / 40 = 3.5\, rounds$.
Distance covered $= 3.5 \times (2 \pi r) = 3.5 \times 2 \times (22/7) \times 100 = 7 \times (22/7) \times 100 = 2200\, m$.
After $3.5\, rounds$, the athlete will be at the diametrically opposite point of the starting position.
Displacement $= \text{Diameter of the circular track} = 200\, m$.

(B) The radius of the circular orbit is $r = 42250 \, km = 42250000 \, m$.
The time taken to complete one revolution is $t = 24 \, h = 24 \times 60 \times 60 \, s = 86400 \, s$.
The speed $v$ of an object moving in a circular path is given by the formula $v = \frac{2 \pi r}{t}$.
Substituting the values: $v = \frac{2 \times (22/7) \times 42250000}{86400}$.
$v = \frac{2 \times 22 \times 42250000}{7 \times 86400} \approx 3073.74 \, m s^{-1}$.
Rounding to the nearest whole number,the speed is approximately $3074 \, m s^{-1}$.

(A) Displacement is defined as the shortest distance between the initial and final positions of an object.
In a circular path of radius $r$,after completing half a circle,the particle moves from one end of the diameter to the other end.
The initial position and final position are diametrically opposite.
The shortest distance between these two points is the diameter of the circle.
Therefore,displacement = $2r$.

(B) boy on a merry-go-round is performing circular motion.
In circular motion,even if the speed remains constant,the direction of motion changes at every point.
Since velocity is a vector quantity (having both magnitude and direction),a change in direction results in a change in velocity.
Therefore,the boy is in accelerated motion.

(B) Displacement is defined as the shortest distance between the initial and final positions of an object.
When a body completes half a revolution on a circular path of radius $R$,it moves from one end of the diameter to the other end.
The initial position is at one end of the diameter,and the final position is at the diametrically opposite end.
The shortest distance between these two points is the diameter of the circle.
Therefore,the displacement is equal to the diameter,which is $2 \times R = 2\,R$.

(A) The speed of an object moving in a circular path is defined as the distance traveled per unit time.
For one complete round of a circular track with radius $r$,the distance traveled is equal to the circumference of the circle,which is $2 \pi r$.
Given that the time taken to complete this round is $t$,the speed $v$ is calculated as:
$v = \frac{\text{Distance}}{\text{Time}} = \frac{2 \pi r}{t}$.

(A) Yes. When a body moves along a circular path with a constant speed,its direction of motion changes at every point. Since velocity is a vector quantity (having both magnitude and direction),a change in direction results in a change in velocity. This change in velocity constitutes acceleration,known as centripetal acceleration.

(N/A) force acting towards the centre of a circular path,known as centripetal force,in which a body moves with constant speed,merely changes the direction of motion of the body without changing its speed.

(D) When an object is in circular motion,it experiences a centripetal force directed towards the center,which keeps it on the circular path.
At any given instant,the velocity of the object is directed along the tangent to the circular path.
When the string breaks,the centripetal force is no longer acting on the stone.
According to Newton's first law of motion,the object will continue to move in the direction of its velocity at that instant.
Therefore,the stone will move along a straight line tangential to the circular path at the point where the string broke.

(A) Yes,it is possible for an object to be in a state of accelerated motion while the work done by the force is zero.
Work done $(W)$ is defined as $W = F \cdot s \cdot \cos(\theta)$,where $F$ is the force,$s$ is the displacement,and $\theta$ is the angle between the force and the displacement.
In uniform circular motion,a centripetal force acts on the object directed towards the center of the circle,which causes centripetal acceleration.
Since the displacement of the object at any instant is along the tangent to the circular path and the centripetal force is directed towards the center,the angle $\theta$ between the force and the displacement is $90^{\circ}$.
Since $\cos(90^{\circ}) = 0$,the work done $W = F \cdot s \cdot 0 = 0$. Thus,despite the presence of an external force causing acceleration,the work done is zero.

(B) When the magnitude of acceleration is constant but its direction is always perpendicular to the velocity of the particle,the particle undergoes uniform circular motion.
In this case,the acceleration acts as a centripetal acceleration,which changes the direction of the velocity vector continuously while keeping its magnitude constant.
Therefore,the path followed by the particle is a circular path.

(N/A) An object moving in a circular path with a constant speed is an example of an accelerated body. This is because the direction of motion changes at every point along the circular path,which results in a change in velocity. Since acceleration is defined as the rate of change of velocity,the body is considered to be accelerated. $A$ common example is a dust particle on the circumference of a revolving bicycle wheel moving at a constant speed.

(B) An athlete moving along a circular path is considered to be in accelerated motion because the direction of the athlete's velocity changes at every point along the path.
Even if the speed (magnitude of velocity) remains constant,the change in direction implies a change in velocity.
Since acceleration is defined as the rate of change of velocity,a continuous change in the direction of motion results in continuous acceleration.

(B) When the acceleration of a particle is constant in magnitude but its direction is always perpendicular to the velocity of the particle,the particle follows a circular path. This is the characteristic condition for uniform circular motion,where the centripetal acceleration changes the direction of velocity continuously while keeping the speed constant.

(N/A) Uniform circular motion is defined as the motion of an object traveling along a circular path at a constant speed. In this type of motion,although the speed remains constant,the direction of velocity changes continuously at every point along the path,which means the object is constantly accelerating.

(N/A) In circular motion,the direction of motion changes at each and every point along the path because the velocity vector is always tangent to the circle at any given point.
$(b)$ Displacement is the shortest distance between the initial and final positions. If a person starts at point $A$ and covers half of the circular path to reach point $B$,the displacement is the straight-line distance between $A$ and $B$,which is equal to the diameter of the circle $(2r)$.
Diagram: The image shows a circle with points $A$ and $B$ at opposite ends of a diameter,with the radius $r$ indicated from the center to point $B$. The displacement is the straight line $AB = 2r$.

(N/A) Uniform circular motion is an example of motion where the speed of the body remains constant,but its velocity changes continuously due to the change in the direction of motion at every point along the circular path.
The diagram below illustrates this motion,where $v$ represents the velocity vector at different points on the circular path. Since the direction of the tangent changes at every point,the velocity vector also changes,even though the magnitude of velocity (speed) remains constant.

(N/A) Acceleration is defined as the rate of change of velocity of an object with respect to time. Mathematically, it is the ratio of the change in velocity to the time taken for that change: $a = \frac{v - u}{t}$.
$(b)$ When a stone moves in a circular path with a constant speed, its direction of motion changes at every point. Since velocity is a vector quantity (speed with direction), the change in direction implies a change in velocity. This type of motion is known as $\text{Uniform circular motion}$.

(N/A) The force that keeps an object moving in a circular path with constant acceleration is known as the centripetal force.
$(b)$ If this force is absent,the object will no longer move in a circular path. Instead,it will move in a straight line along the direction of the tangent to the circular path at the point where the force ceased to act,in accordance with Newton's first law of motion.

(C) Work done is defined by the formula $W = F S \cos \theta$,where $F$ is the force,$S$ is the displacement,and $\theta$ is the angle between the force and displacement vectors.
When a body completes one full revolution in a circular path,the final position coincides with the initial position,making the net displacement $S = 0$.
Since the centripetal force acts perpendicular to the direction of motion (tangential velocity) at every point,the work done by the centripetal force is zero.
Therefore,the total work done by the force when a body moves in a circular path is $W = F \times 0 = 0$.

(N/A) Uniform circular motion is defined as the motion of an object traveling along a circular path at a constant speed.
$(b)$ Yes,uniform circular motion is an accelerated motion. Even though the speed of the object remains constant,the direction of its velocity changes continuously at every point along the circular path. Since acceleration is defined as the rate of change of velocity,and velocity is a vector quantity (having both magnitude and direction),the continuous change in direction results in a continuous change in velocity,thereby causing acceleration.

(A) When an object completes half of a circular path, the distance covered is equal to half of the circumference of the circle.
Distance $= \pi R = \frac{22}{7} \times 7 = 22 \, m$.
The displacement is the shortest distance between the initial and final positions, which is the diameter of the circle.
Displacement $= 2R = 2 \times 7 = 14 \, m$.
The object possesses uniform circular motion because it moves with a constant speed along a circular path.

(N/A) Distance: It is defined as the total path length covered by a body.
Displacement: It is defined as the shortest distance between the initial and final positions of a body.
Given that the circumference of the circular park is $176 \, m$. The body completes one revolution in $4$ minutes.
After $6$ minutes,the number of revolutions completed $= \frac{6}{4} = 1.5$ revolutions.
Since $1$ revolution brings the body back to the starting point,after $1.5$ revolutions,the body will be at a position diametrically opposite to the starting point.
Given $2 \pi r = 176 \, m$,where $r$ is the radius.
$r = \frac{176}{2 \pi} = \frac{176 \times 7}{2 \times 22} = 28 \, m$.
The displacement after $1.5$ revolutions is the diameter of the circular path.
Displacement $= 2r = 2 \times 28 = 56 \, m$.

$(a)$ When a body moves along a circular path, it is said to be in circular motion.
$(b)$ Uniform circular motion is considered an accelerated motion because the direction of the velocity vector changes at every point along the path, even though the magnitude (speed) remains constant. Since acceleration is defined as the rate of change of velocity, a change in direction constitutes acceleration.
$(c)$ The distance covered in one revolution is the circumference of the circular orbit, given by $2 \pi r$.
Distance $= 2 \times 3.14 \times 42250 \, km = 265330 \, km$.
Time taken $(t) = 24 \, h$.
Speed $= \frac{\text{Distance}}{\text{Time}} = \frac{265330 \, km}{24 \, h} \approx 11055.42 \, km/h$.

(N/A) In uniform linear motion,a body moves along a straight line with constant speed. In uniform circular motion,a body moves along a circular path with constant speed.
$(b)$ $(i)$ Motion of a point on the rim of a rotating wheel.
$(ii)$ Motion of a satellite around a planet.
$(iii)$ Motion of the Moon around the Earth.
$(iv)$ Motion of the tip of the second hand of a clock.
$(c)$ Yes,uniform circular motion is an accelerated motion. Although the speed remains constant,the direction of motion changes continuously at every point along the circular path. Since velocity is a vector quantity (speed with direction),a change in direction implies a change in velocity,which constitutes acceleration.

(A) The diameter of the circular track is $D = 105 \, m$. Therefore, the radius $R = \frac{105}{2} = 52.5 \, m$.
The time taken to complete one round is $t = 5 \, minutes = 5 \times 60 \, s = 300 \, s$.
The distance covered in one round is equal to the circumference of the circle, which is $2 \pi R$.
Distance $= 2 \times 3.14 \times 52.5 = 329.7 \, m$.
Speed $= \frac{\text{Distance}}{\text{Time}} = \frac{329.7}{300} = 1.099 \, m/s \approx 1.1 \, m/s$.

(A) The time taken for one revolution is $t = 1 \text{ hour } 30 \text{ minutes} = 90 \text{ minutes}$.
Converting this into seconds: $t = 90 \times 60 = 5400 \text{ s}$.
The angle covered in one complete revolution is $\theta = 2\pi \text{ radians}$.
The formula for angular speed is $\omega = \frac{\theta}{t}$.
Substituting the values: $\omega = \frac{2\pi}{5400} = \frac{\pi}{2700} \text{ rad/s}$.
Calculating the numerical value: $\omega \approx \frac{3.14159}{2700} \approx 1.1635 \times 10^{-3} \text{ rad/s}$.
Thus,the angular speed is approximately $1.16 \times 10^{-3} \text{ rad/s}$.

(A) Given:
Radius $r = 42,250 \, km = 42,250,000 \, m$
Time period $T = 24 \, \text{hours} = 24 \times 60 \times 60 \, s = 86,400 \, s$
The linear velocity $v$ of an object moving in a circular path is given by the formula:
$v = \frac{2 \pi r}{T}$
Substituting the values:
$v = \frac{2 \times 3.14159 \times 42,250 \, km}{86,400 \, s}$
$v \approx \frac{265,464}{86,400} \, km/s$
$v \approx 3.07 \, km/s$
Thus, the linear velocity of the satellite is approximately $3.07 \, km/s$.

(DISTANCE = 231 M, DISPLACEMENT = 49 M) Given: Diameter $d = 49 \,m$,Time for one round $T = 20 \,s$,Total time $t = 30 \,s$.
First,calculate the number of rounds completed in $30 \,s$:
$n = \frac{t}{T} = \frac{30}{20} = 1.5 \, \text{rounds}$.
Distance covered is the total path length:
$S = n \times (2 \pi r) = n \times (\pi d) = 1.5 \times \frac{22}{7} \times 49 = 1.5 \times 22 \times 7 = 231 \,m$.
Displacement is the shortest distance between the initial and final positions:
After $1.5$ rounds,the athlete is at the point diametrically opposite to the starting point.
Therefore,the displacement is equal to the diameter of the circular track,which is $49 \,m$.

(D) When an object moves in a circular path at a constant speed,its velocity changes continuously due to the change in direction at every point.
This change in velocity gives rise to an acceleration known as centripetal acceleration.
By definition,centripetal acceleration is always directed towards the centre of the circular path.
Therefore,the correct option is $D$.

(A) The statement is $True$.
Work done $(W)$ is defined by the formula $W = F \cdot s \cdot \cos(\theta)$,where $F$ is the force,$s$ is the displacement,and $\theta$ is the angle between the force and the displacement.
In circular motion,the centripetal force acts towards the center of the circle,while the displacement (tangential velocity) is always perpendicular to the radius.
Since the angle $\theta$ between the centripetal force and the displacement is $90^{\circ}$,the work done is $W = F \cdot s \cdot \cos(90^{\circ})$.
Because $\cos(90^{\circ}) = 0$,the work done by the centripetal force is $0$.