Curvilinear Motion MCQ Quiz in தமிழ் - Objective Question with Answer for Curvilinear Motion - இலவச PDF ஐப் பதிவிறக்கவும்
Last updated on Mar 11, 2025
Latest Curvilinear Motion MCQ Objective Questions
Top Curvilinear Motion MCQ Objective Questions
Curvilinear Motion Question 1:
A car travels on a horizontal circular track of radius 9 m, starting from rest at a constant tangential acceleration of 3 m/s2. What is the resultant acceleration of the car, 2 sec after starting?
Answer (Detailed Solution Below)
Curvilinear Motion Question 1 Detailed Solution
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 \(a = \sqrt {\alpha _t^2 + \alpha _n^2}\)
\(= \sqrt {{3^2} + {4^2}} = 5\;m/{s^2}\)
Curvilinear Motion Question 2:
The radial component of velocity and acceleration in curvilinear motion are
Answer (Detailed Solution Below)
Curvilinear Motion Question 2 Detailed Solution
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:
- Radial components denoted by \(\overrightarrow{r}\)
- 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\(\overrightarrow{r}~=~\overrightarrow{PA_1}~=~\overrightarrow{PA_2}~+~\overrightarrow{A_2A_1}~=~(Cos\theta)i~+~(Sin\theta)j\)
\(\overrightarrow{\theta}~=~\overrightarrow{PB_1}~=~\overrightarrow{PB_2}~+~\overrightarrow{B_2B_1}~=~-(Sin\theta)i~+~(Cos\theta)j\)
Velocity Component: When the above identities are differentiated with respect to time t.\(\frac{\mathrm{d} }{\mathrm{d} t}(\hat{r})~=~[-(Sin\theta)i~+~(Cos\theta)j]\frac{d\theta}{dt}~=~\dot{\theta}\hat{\theta}\)\(\frac{\mathrm{d} }{\mathrm{d} t}(\hat{\theta})~=~[(-\cos \theta)i~+~(-sin\theta)j]\frac{d\theta}{dt}~=~-\dot{\theta ~\hat{r}}\)
Denoting \(\left|\overrightarrow{r} \right|~as~r\), we can write \(\overrightarrow{r}~=~r~\hat{r}\)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,
\(\overrightarrow{V}~=~\frac{dr}{dt}(\hat{r})~+~r(\dot{\theta}~\hat{\theta})~=~\dot{r~\hat{r}}~+~r~({\dot{\theta} ~\hat{\theta}})\)
Radial component, \(V_r~=~\dot{r}~=~\frac{dr}{dt}\)Transverse component, \(V_{\theta}~=~r~\dot{\theta}~=~r\frac{d\theta}{dt}\)
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.
\(\overrightarrow{a}~=~\ddot{r}~\hat{r}~+~\dot{r}(\dot{\theta~\hat{\theta}})~+~(r~\frac{d\dot{\theta}}{dt}~+~\dot{\theta}~\frac{dr}{dt})~\hat{\theta}~+~r\dot{\theta}~(-\dot{\theta~\hat{r}})~=~(\ddot{r}~-~r{\dot{\theta}^2})\hat{r}~+~(r\ddot{\theta}~+~2\dot{r}\dot{\theta})\hat{\theta}\)
The radial component of acceleration,\(a_r~=~\ddot{r}~-~r{\dot{\theta}}^2\)
Curvilinear Motion Question 3:
A body moves with a speed of 10 m/s in the curved path of 25 m radius of curvature. If the tangential acceleration is 3 m/s2, then total acceleration for the body will be:
Answer (Detailed Solution Below)
Curvilinear Motion Question 3 Detailed Solution
CONCEPT:
Centripetal Acceleration (ac):
- 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 (at):
- 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 at = 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 (ac):
\(\therefore {a_c} = \frac{{{v^2}}}{r}\)
\( \Rightarrow {a_c} = \frac{{{{\left( {10} \right)}^2}}}{{25}} = \frac{{100}}{{25}} = 4\;m/{s^2}\)
Hence, net acceleration
\(a = \sqrt {a_t^2 + a_c^2} = \sqrt {{4^2} + {3^2}} = 5\;m/{s^2}\)Curvilinear Motion Question 4:
A disc arranged in a vertical plane has two groves of same length directed along the vertical chord AB and CD as shown in the fig. The same particles slide down along AB and CD. The ratio of the time tAB/tCD is:
Answer (Detailed Solution Below)
Curvilinear Motion Question 4 Detailed Solution
\(\begin{array}{l} {S_{AB}} = \frac{1}{2}gt_{AB}^2\\ {S_{CD}} = \frac{1}{2}g\cos 60^\circ t_{CD}^2\\ {\rm{But}},{S_{AB}} = {S_{CD}}\\ \therefore \frac{{{S_{AB}}}}{{{S_{CD}}}} = \frac{{\frac{1}{2}g\;t_{AB}^2}}{{\frac{1}{2}g\cos 60^\circ t_{CD}^2}}\\ {\rm{Or}},{\rm{\;}}1 = 2\frac{{t_{AB}^2}}{{t_{CD}^2}} \Rightarrow \frac{{{t_{AB}}}}{{{t_{CD}}}} = 1:\sqrt 2 \end{array}\)
Curvilinear Motion Question 5:
If a wheel of radius R rolls without slipping through an angle θ, what is the relationship between the distance the wheel rolls, x, and the product Rθ?
Answer (Detailed Solution Below)
Curvilinear Motion Question 5 Detailed Solution
Since the wheel is in constant contact with the ground, the length of the arc correlating to the angle is also equal to x.
Rolling without slipping condition: x = Rθ.Curvilinear Motion Question 6:
A stone, is tied to one end of a spring 50 cm long, is whirled in a horizontal circle with a constant speed. If the stone makes 10 revolutions in 20 s, what is the magnitude of acceleration of the stone
Answer (Detailed Solution Below)
Curvilinear Motion Question 6 Detailed Solution
Time Period
\(T = \frac{{Total\;Time}}{{No.\;of\;revolutions}} = \frac{{20}}{{10}} = 2\;sec.\)
\(\therefore {a_c} = \frac{{4{\pi ^2}}}{{{T^2}}}.r = \frac{{4{\pi ^2}}}{{{2^2}}} \times \left( {50\times10^{-2}} \right) = 4.93\frac{m}{{{s^2}}} = 493\;cm/{s^2}\)
Curvilinear Motion Question 7:
A stone is thrown from as elevation of 2m such that it clears a wall 8m high situated at a horizontal distance of 35m. If the angle of projection is 60° with respect to the horizontal, What should be the minimum velocity of projection?
Answer (Detailed Solution Below)
20.97 m/s
Curvilinear Motion Question 7 Detailed Solution
Equation of path,
\(y = xtan\theta - \frac{{g{x^2}}}{{2u_0^2{{\cos }^2}\theta }}\)
θ = 60°, x = 35m, y = 8 -2 = 6m
⇒ u0 = 20.97 m/s
Curvilinear Motion Question 8:
A particle is projected freely into the air at an angle θ from the ground at and velocity u0 as shown in given figure. Neglect the air resistance. Then equation of the path of motion of the particle is given by
Answer (Detailed Solution Below)
\(xtan\theta - \frac{{g{x^2}}}{{2u_0^2co{s^2}\theta }}\)
Curvilinear Motion Question 8 Detailed Solution
U0x = u0cosθ& u0y = u0sinθ
X = u0 t = (u0cosθ )t
\(\Rightarrow t = \frac{x}{{{u_0}cos\theta }}\) -----------(1)
Y = u0y t – ½ gt2 --------- (2)
Substitute equation (1) is equation (2)
\(Y = x\;tan\theta \;-\;\frac{1}{2}\frac{{g{x^2}}}{{u_0^2{{\cos }^2}\theta }}\)
Curvilinear Motion Question 9:
A point particle moves along a path y = f(x) then radius of curvature of path at a particular location is given by
Answer (Detailed Solution Below)
\(\frac{{{{\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]}^{\frac{3}{2}}}}}{{\frac{{{d^2}y}}{{d{x^2}}}}}\)
Curvilinear Motion Question 9 Detailed Solution
Explanation:
Radius of Curvature:
We can draw a circle that closely fits nearby points on a local section of a curve, as follows
- We say the curve and the circle osculate since the 2 curves have the same tangent and curvature at the point where they meet.
- The radius of curvature of the curve at a particular point is defined as the radius of the approximating circle. This radius changes as we move along the curve.
- The formula for the radius of curvature at any point x for the curve y = f(x) is given by:
Radius of curvature = \(\frac{{{{\left[ {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} \right]}^{\frac{3}{2}}}}}{{\frac{{{d^2}y}}{{d{x^2}}}}}\)
So, option (3) is the correct answer.
Curvilinear Motion Question 10:
A point particle moves along a path y = 4x2 with a constant speed of 10 m/s. Then tangential components of acceleration at x = 3 m is
Answer (Detailed Solution Below)
Zero
Curvilinear Motion Question 10 Detailed Solution
Tangential component of acceleration is responsible for change in magnitude of velocity. If magnitude of velocity is constant then at = O.