Curvilinear Motion MCQ Quiz - Objective Question with Answer for Curvilinear Motion - Download Free PDF

Concept:

To find the time of meeting of the two shots, we consider the relative motion of the projectiles in the horizontal direction.

Given:

Distance between guns, d = 40 m

Velocity of shot from Gun A, v_A = 350 m/s

Velocity of shot from Gun B, v_B = 300 m/s

Angle of projection, \(\theta = 30^\circ\)

Calculation:

Resolving velocities into horizontal components:

\( v_{Ax} = v_A \cos 30^\circ = 350 \times \frac{\sqrt{3}}{2} = 175\sqrt{3} \) m/s

\( v_{Bx} = v_B \cos 30^\circ = 300 \times \frac{\sqrt{3}}{2} = 150\sqrt{3} \) m/s

The relative velocity in the horizontal direction is:

\( v_{\text{relative}, x} = v_{Ax} + v_{Bx} = 175\sqrt{3} + 150\sqrt{3} = 325\sqrt{3} \) m/s

Using the formula,

\( \text{Time} = \frac{\text{Distance}}{\text{Relative Velocity}} \)

\( t = \frac{40}{325\sqrt{3}} \)

Approximating \(\sqrt{3} \approx 1.732\) ,

\( t = \frac{40}{325 \times 1.732} = \frac{40}{562.9} \approx \frac{4}{65}\) seconds

Explanation:

Projectile Motion Under Gravity

A projectile, when launched under the influence of gravity alone, follows a specific path. This path is influenced by the initial velocity, angle of projection, and the acceleration due to gravity. The trajectory of a projectile is described as:

• Parabolic Path: The projectile follows a curved path known as a parabola. This is because the horizontal motion is uniform while the vertical motion is uniformly accelerated.

Analyzing the Given Options

1. "Exaggerating" (Incorrect)

   • The term "exaggerating" does not relate to the path of a projectile. It describes an action or statement that is overstated or made to seem more extreme than it actually is.

2. "Parabola" (Correct)

   • Under the influence of gravity alone, a projectile follows a parabolic trajectory. This is due to the combination of uniform horizontal motion and uniformly accelerated vertical motion.

3. "Straight Line" (Incorrect)

   • If no external forces like gravity acted on the projectile, it would move in a straight line. However, gravity causes the projectile to follow a curved path.

4. "Circle" (Incorrect)

   • A circular path would imply a centripetal force acting perpendicular to the motion, which is not the case for a projectile influenced only by gravity.

Tangential acceleration = r × angular acceleration

\(\begin{array}{l} {\alpha _t} = r\alpha \\ \Rightarrow \alpha = \frac{3}{9} = \frac{1}{3}rad/{s^2} \end{array}\)

Now,

\(\begin{array}{l} \omega = \omega_o + \alpha t\\ {\omega _o} = 0\\ \Rightarrow \omega = \frac{1}{3} \times 2 = \frac{2}{3}rad/s \end{array}\)

Now, Normal acceleration \(\left( {{\alpha _n}} \right) = r{\omega ^2}\)

\(= 9 \times \frac{4}{9} = 4\;m/{s^2}\)

∴ Resultant acceleration \(a = \sqrt {\alpha _t^2 + \alpha _n^2}\)

\(= \sqrt {{3^2} + {4^2}} = 5\;m/{s^2}\)

Explanation:

Explanation:

Radial and Transverse co-ordinates:

• In this system, the position of the particle is defined by the polar coordinates r and θ
• Further, the velocity and acceleration of the particle are resolved into components along and perpendicular to the position vector \(\overrightarrow{r}\). These components are called:

1. Radial components denoted by \(\overrightarrow{r}\)
2. Transverse components denoted by \(\overrightarrow{θ}\)

The unit vectors of \(\overrightarrow{r}\) and \(\overrightarrow{θ}\) are represented by \(\hat{r}\) and \(\hat{θ}\) respectively. Here A₁A₂ and B₁B₂ are perpendicular to the x-axis and are drawn from the endpoints A₁ and B₁ of the unit vectors \(\hat{r}\) and \(\hat{θ}\) respectivel

 
Furthur: PA₁ = \(\left|\hat{r} \right| = 1\) and PB₂ = \(\left|\hat{θ} \right| = 1\)
That gives: A₁A₂ = Sin θ and B₁B₂ = Cos θ, PA₂ = Cos θ and PB₂ = Sin θ
Applying the triangle law of addition of vectors\(\overrightarrow{r}~=~\overrightarrow{PA_1}~=~\overrightarrow{PA_2}~+~\overrightarrow{A_2A_1}~=~(Cos\theta)i~+~(Sin\theta)j\)

\(\overrightarrow{\theta}~=~\overrightarrow{PB_1}~=~\overrightarrow{PB_2}~+~\overrightarrow{B_2B_1}~=~-(Sin\theta)i~+~(Cos\theta)j\)

\(\frac{\mathrm{d} }{\mathrm{d} t}(\hat{\theta})~=~[(-\cos \theta)i~+~(-sin\theta)j]\frac{d\theta}{dt}~=~-\dot{\theta ~\hat{r}}\)

In terms of the position vector \(\overrightarrow{r}\), the velocity vector is defined as

\(\overrightarrow{V}~=~\frac{d}{dt}(\overrightarrow{r})~=~\frac{d}{dt}(r~\hat{r})~=~\frac{dr}{dt} \hat{r}~+~r~\frac{d\hat{r}}{dt}\)
Replacing \(\frac{d\hat{r}}{dt}~by~(\dot{\theta}~\hat{\theta})\) as worked out above,

\(\overrightarrow{V}~=~\frac{dr}{dt}(\hat{r})~+~r(\dot{\theta}~\hat{\theta})~=~\dot{r~\hat{r}}~+~r~({\dot{\theta} ~\hat{\theta}})\)

Acceleration:

\(\overrightarrow{a}~=~\frac{d}{dt}(\overrightarrow{V})~=~\frac{d}{dt}[\dot{r}~\hat{r}~+~(r\dot{\theta})\hat{\theta}]\\ ~~~~~=~\frac{d\dot{r}}{dt}~\hat{r}~+~\dot{r}\frac{d\hat{r}}{dt}~+~\frac{d(r\dot{\theta})}{dt}~\hat{\theta}~+~r~{\dot{\theta}~\frac{d\hat{\theta}}{dt}} \\~~~~~=~\frac{d\dot{r}}{dt}~\hat{r}~+~\dot{r}\frac{d\hat{r}}{dt}~+~(r\frac{d\dot{\theta}}{dt}~+~\dot{\theta}~\frac{dr}{dt})\hat{\theta}+~r~{\dot{\theta}~\frac{d\hat{\theta}}{dt}}\)

Substituting the values for \(\frac{d}{dt}(\hat{r})\) and \(\frac{d}{dt}(\hat{\theta})\) in acceleration.

\(\overrightarrow{a}~=~\ddot{r}~\hat{r}~+~\dot{r}(\dot{\theta~\hat{\theta}})~+~(r~\frac{d\dot{\theta}}{dt}~+~\dot{\theta}~\frac{dr}{dt})~\hat{\theta}~+~r\dot{\theta}~(-\dot{\theta~\hat{r}})~=~(\ddot{r}~-~r{\dot{\theta}^2})\hat{r}~+~(r\ddot{\theta}~+~2\dot{r}\dot{\theta})\hat{\theta}\)

\(a_r~=~\ddot{r}~-~r{\dot{\theta}}^2\)

CONCEPT:

Centripetal Acceleration (a_c): 

• Acceleration acting on the object undergoing uniform circular motion is called centripetal acceleration.
• It always acts on the object along the radius towards the center of the circular path.
• The magnitude of centripetal acceleration,

\(a = \frac{{{v^2}}}{r}\)

Where v = velocity of the object and r = radius

Tangential acceleration (a_t):

• It acts along the tangent to the circular path in the plane of the circular path.
• Mathematically Tangential acceleration is written as

\(\overrightarrow {{a_t}} = \vec \alpha \times \vec r \)

Where α = angular acceleration and r = radius

CALCULATION:

Given – v = 10 m/s, r = 25 m and a_t = 3 m/s²

• Net acceleration is the resultant acceleration of centripetal acceleration and tangential acceleration i.e.,

\(a = \sqrt {a_c^2 + a_t^2} \)

Centripetal Acceleration (a_c):

\(\therefore {a_c} = \frac{{{v^2}}}{r}\)

\( \Rightarrow {a_c} = \frac{{{{\left( {10} \right)}^2}}}{{25}} = \frac{{100}}{{25}} = 4\;m/{s^2}\)

Hence, net acceleration 

CONCEPT:

Centripetal Acceleration (a_c): 

• Acceleration acting on the object undergoing uniform circular motion is called centripetal acceleration.
• It always acts on the object along the radius towards the center of the circular path.
• The magnitude of centripetal acceleration,

\(a = \frac{{{v^2}}}{r}\)

Where v = velocity of the object and r = radius

Tangential acceleration (a_t):

• It acts along the tangent to the circular path in the plane of the circular path.
• Mathematically Tangential acceleration is written as

\(\overrightarrow {{a_t}} = \vec \alpha \times \vec r \)

Where α = angular acceleration and r = radius

CALCULATION:

Given – v = 10 m/s, r = 25 m and a_t = 3 m/s²

• Net acceleration is the resultant acceleration of centripetal acceleration and tangential acceleration i.e.,

\(a = \sqrt {a_c^2 + a_t^2} \)

Centripetal Acceleration (a_c):

\(\therefore {a_c} = \frac{{{v^2}}}{r}\)

\( \Rightarrow {a_c} = \frac{{{{\left( {10} \right)}^2}}}{{25}} = \frac{{100}}{{25}} = 4\;m/{s^2}\)

Hence, net acceleration 

Tangential acceleration = r × angular acceleration

\(\begin{array}{l} {\alpha _t} = r\alpha \\ \Rightarrow \alpha = \frac{3}{9} = \frac{1}{3}rad/{s^2} \end{array}\)

Now,

\(\begin{array}{l} \omega = \omega_o + \alpha t\\ {\omega _o} = 0\\ \Rightarrow \omega = \frac{1}{3} \times 2 = \frac{2}{3}rad/s \end{array}\)

Now,

Normal acceleration \(\left( {{\alpha _n}} \right) = r{\omega ^2}\)

\(= 9 \times \frac{4}{9} = 4\;m/{s^2}\)

∴ Resultant acceleration \(a = \sqrt {\alpha _t^2 + \alpha _n^2}\)

\(= \sqrt {{3^2} + {4^2}} = 5\;m/{s^2}\)

Distance \(= \frac{1}{4} \times circumference\) 

\(\begin{array}{l} = \frac{1}{4} \times2 \pi(2)= \pi \\ \Rightarrow time\ taken = \frac{{distance}}{{speed}} = \frac{\pi }{{1.57}} = 2sec \end{array}\)

Explanation:

Explanation:

Radial and Transverse co-ordinates:

• In this system, the position of the particle is defined by the polar coordinates r and θ
• Further, the velocity and acceleration of the particle are resolved into components along and perpendicular to the position vector \(\overrightarrow{r}\). These components are called:

1. Radial components denoted by \(\overrightarrow{r}\)
2. Transverse components denoted by \(\overrightarrow{θ}\)

The unit vectors of \(\overrightarrow{r}\) and \(\overrightarrow{θ}\) are represented by \(\hat{r}\) and \(\hat{θ}\) respectively. Here A₁A₂ and B₁B₂ are perpendicular to the x-axis and are drawn from the endpoints A₁ and B₁ of the unit vectors \(\hat{r}\) and \(\hat{θ}\) respectivel

 
Furthur: PA₁ = \(\left|\hat{r} \right| = 1\) and PB₂ = \(\left|\hat{θ} \right| = 1\)
That gives: A₁A₂ = Sin θ and B₁B₂ = Cos θ, PA₂ = Cos θ and PB₂ = Sin θ
Applying the triangle law of addition of vectors\(\overrightarrow{r}~=~\overrightarrow{PA_1}~=~\overrightarrow{PA_2}~+~\overrightarrow{A_2A_1}~=~(Cos\theta)i~+~(Sin\theta)j\)

\(\overrightarrow{\theta}~=~\overrightarrow{PB_1}~=~\overrightarrow{PB_2}~+~\overrightarrow{B_2B_1}~=~-(Sin\theta)i~+~(Cos\theta)j\)

\(\frac{\mathrm{d} }{\mathrm{d} t}(\hat{\theta})~=~[(-\cos \theta)i~+~(-sin\theta)j]\frac{d\theta}{dt}~=~-\dot{\theta ~\hat{r}}\)

In terms of the position vector \(\overrightarrow{r}\), the velocity vector is defined as

\(\overrightarrow{V}~=~\frac{d}{dt}(\overrightarrow{r})~=~\frac{d}{dt}(r~\hat{r})~=~\frac{dr}{dt} \hat{r}~+~r~\frac{d\hat{r}}{dt}\)
Replacing \(\frac{d\hat{r}}{dt}~by~(\dot{\theta}~\hat{\theta})\) as worked out above,

\(\overrightarrow{V}~=~\frac{dr}{dt}(\hat{r})~+~r(\dot{\theta}~\hat{\theta})~=~\dot{r~\hat{r}}~+~r~({\dot{\theta} ~\hat{\theta}})\)

Acceleration:

\(\overrightarrow{a}~=~\frac{d}{dt}(\overrightarrow{V})~=~\frac{d}{dt}[\dot{r}~\hat{r}~+~(r\dot{\theta})\hat{\theta}]\\ ~~~~~=~\frac{d\dot{r}}{dt}~\hat{r}~+~\dot{r}\frac{d\hat{r}}{dt}~+~\frac{d(r\dot{\theta})}{dt}~\hat{\theta}~+~r~{\dot{\theta}~\frac{d\hat{\theta}}{dt}} \\~~~~~=~\frac{d\dot{r}}{dt}~\hat{r}~+~\dot{r}\frac{d\hat{r}}{dt}~+~(r\frac{d\dot{\theta}}{dt}~+~\dot{\theta}~\frac{dr}{dt})\hat{\theta}+~r~{\dot{\theta}~\frac{d\hat{\theta}}{dt}}\)

Substituting the values for \(\frac{d}{dt}(\hat{r})\) and \(\frac{d}{dt}(\hat{\theta})\) in acceleration.

\(\overrightarrow{a}~=~\ddot{r}~\hat{r}~+~\dot{r}(\dot{\theta~\hat{\theta}})~+~(r~\frac{d\dot{\theta}}{dt}~+~\dot{\theta}~\frac{dr}{dt})~\hat{\theta}~+~r\dot{\theta}~(-\dot{\theta~\hat{r}})~=~(\ddot{r}~-~r{\dot{\theta}^2})\hat{r}~+~(r\ddot{\theta}~+~2\dot{r}\dot{\theta})\hat{\theta}\)

\(a_r~=~\ddot{r}~-~r{\dot{\theta}}^2\)

Concept:

Angular acceleration is the rate of change of angular velocity

\(\alpha = \frac{{\mathop \omega \nolimits_2 - \mathop \omega \nolimits_1 }}{t}\)

Calculation:

Given

\(\begin{array}{l} \mathop \omega \nolimits_2 = 800~rpm = \frac{{800 \times 2\pi }}{{60}}\frac{{rad}}{s}\\ \mathop \omega \nolimits_1 = 300~rpm= \frac{{300 \times 2\pi }}{{60}}\frac{{rad}}{s} \end{array}\)

t = 10 sec

\(\alpha = \frac{{\mathop \omega \nolimits_2 - \mathop \omega \nolimits_1 }}{t}\)

\(\alpha =\frac{2\pi}{60} \times \frac{{800 - 300}}{10}\)

\(\alpha = 5.24\frac{{rad}}{{\mathop s\nolimits^2 }}\)

Concept:

Distance covered by translation, \({\rm{S\;}} = {\rm{\;u}} \times {\rm{t\;}} + {\rm{\;}}\frac{1}{2}a{t^2}\)

u is the initial motion, t is the time and a is the acceleration

Calculation:

Given u = 0, a = 5 m/s² and t = 5 s

\({\rm{S\;}} = {\rm{\;u}} \times {\rm{t\;}} + {\rm{\;}}\frac{1}{2}a{t^2} = \frac{1}{2} \times 5 \times {5^2} = 62.5\;m\)

Concept:
Curvilinear motion: This type of motion occurs when an object moves in a curved path.

• The velocity of a particle moving in a curved path is always tangential to the path moved. Hence, the tangential component of velocity is equal to velocity itself as the particle moves tangentially and the normal component is zero.
• The normal component is perpendicular to the velocity vector and passes through the center of curvature. This type of coordinate system, in which the motion of particles is described in two directions is known as a normal and tangential coordinate system.
• R is the particle moving in a curved path shown in the diagram.
• Hence, V_{normal component} = 0 and V_{tangential component} = velocity of the particle

The tangential direction of velocity = Velocity of particle

CONCEPT:

Centripetal Acceleration (a_c): 

• Acceleration acting on the object undergoing uniform circular motion is called centripetal acceleration.
• It always acts on the object along the radius towards the center of the circular path.
• The magnitude of centripetal acceleration,

\(a = \frac{{{v^2}}}{r}\)

Where v = velocity of the object and r = radius

Tangential acceleration (a_t):

• It acts along the tangent to the circular path in the plane of the circular path.
• Mathematically Tangential acceleration is written as

\(\overrightarrow {{a_t}} = \vec \alpha \times \vec r \)

Where α = angular acceleration and r = radius

CALCULATION:

Given – v = 10 m/s, r = 25 m and a_t = 3 m/s²

• Net acceleration is the resultant acceleration of centripetal acceleration and tangential acceleration i.e.,

\(a = \sqrt {a_c^2 + a_t^2} \)

Centripetal Acceleration (a_c):

\(\therefore {a_c} = \frac{{{v^2}}}{r}\)

\( \Rightarrow {a_c} = \frac{{{{\left( {10} \right)}^2}}}{{25}} = \frac{{100}}{{25}} = 4\;m/{s^2}\)

Hence, net acceleration 

Concept:

When a body moves along a curved path a net force acts on the body that is directed towards the center of curvature. This force is called the centripetal Force, F_C. 

\(F_C = \frac{ mV^2}{r}\;\)

where, m is body mass, V is magnitude of the velocity, r is radius of curvature.

Calculation:

Given:

Case 1:

The car A is running on a plane bridge as shown -

The normal force F_A = mg 

Case 2:-

The car B is running on a convex bridge as shown - 

Now this is a case of plane circular motion and in this case, the net centripetal force acts towards the center of curvature.

So the normal force

\(\frac{mV^2}{r}=mg-F_B\)

Or, \(F_B=mg-\frac{mV^2}{r}\;\)

Case 3:

The car C is running on a concave bridge as shown - 

In this case, there is also a net centripetal force acting towards the center of curvature.

So the normal force F_C

\(\frac{mV^2}{r}=F_C-mg\)

\(F_C=mg+\frac{mV^2}{r}\;\)

Hence, by looking at all three equations,

F_C > F_A > F_B 

we can say that the contact force F_C is the maximum among all three forces.

Tangential acceleration = r × angular acceleration

\(\begin{array}{l} {\alpha _t} = r\alpha \\ \Rightarrow \alpha = \frac{3}{9} = \frac{1}{3}rad/{s^2} \end{array}\)

Now,

\(\begin{array}{l} \omega = \omega_o + \alpha t\\ {\omega _o} = 0\\ \Rightarrow \omega = \frac{1}{3} \times 2 = \frac{2}{3}rad/s \end{array}\)

Now,

Normal acceleration \(\left( {{\alpha _n}} \right) = r{\omega ^2}\)

\(= 9 \times \frac{4}{9} = 4\;m/{s^2}\)

∴ Resultant acceleration \(a = \sqrt {\alpha _t^2 + \alpha _n^2}\)

\(= \sqrt {{3^2} + {4^2}} = 5\;m/{s^2}\)