Curvilinear Motion MCQ Quiz in தமிழ் - Objective Question with Answer for Curvilinear Motion - இலவச PDF ஐப் பதிவிறக்கவும்

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 

\(\begin{array}{l} {S_{AB}} = \frac{1}{2}gt_{AB}^2\\ {S_{CD}} = \frac{1}{2}g\cos 60^\circ t_{CD}^2\\ {\rm{But}},{S_{AB}} = {S_{CD}}\\ \therefore \frac{{{S_{AB}}}}{{{S_{CD}}}} = \frac{{\frac{1}{2}g\;t_{AB}^2}}{{\frac{1}{2}g\cos 60^\circ t_{CD}^2}}\\ {\rm{Or}},{\rm{\;}}1 = 2\frac{{t_{AB}^2}}{{t_{CD}^2}} \Rightarrow \frac{{{t_{AB}}}}{{{t_{CD}}}} = 1:\sqrt 2 \end{array}\)

Since the wheel is in constant contact with the ground, the length of the arc correlating to the angle is also equal to x.

Time Period

\(T = \frac{{Total\;Time}}{{No.\;of\;revolutions}} = \frac{{20}}{{10}} = 2\;sec.\)

\(\therefore {a_c} = \frac{{4{\pi ^2}}}{{{T^2}}}.r = \frac{{4{\pi ^2}}}{{{2^2}}} \times \left( {50\times10^{-2}} \right) = 4.93\frac{m}{{{s^2}}} = 493\;cm/{s^2}\)

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

U_0x = u₀cosθ& u_0y = u₀sinθ

X = u₀ t = (u₀cosθ )t

\(\Rightarrow t = \frac{x}{{{u_0}cos\theta }}\) -----------(1)

Y = u_0y t – ½ gt² --------- (2)

Substitute equation (1) is equation (2)

\(Y = x\;tan\theta \;-\;\frac{1}{2}\frac{{g{x^2}}}{{u_0^2{{\cos }^2}\theta }}\)

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.

Tangential component of acceleration is responsible for change in magnitude of velocity. If magnitude of velocity is constant then a_t = O.