Notes-Part-2
Topics to be Learn : Part-2
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Motion in a plane :
Motion in Two Dimensions-Motion in a Plane:
Consider a two—dimensional motion in which P(x1, y1) and Q(x2, y2) are the positions of a particle at instants t1 and t2 respectively (Fig.).
The motion of the particle is, in general, along a curved path, shown by the dashed line in the figure.
Path length : The length of this curved path between P and Q is the distance travelled by the particle.
Displacement : Suppose \(\vec{r}_1=x_1\hat{i}+y_1\hat{j}\) is the position vector of a particle at an initial time t1, and \(\vec{r}_2=x_2\hat{i}+y_2\hat{j}\) is the position vector at the final time t2. Then, the displacement \(\vec{s}\) of the particle during that time interval Δt = t2 — t1 is
\(\vec{s}=Δ\vec{r}=\vec{r}_2-\vec{r}_1=(x_2-x_1)\hat{i}+(y_2-y_1)\hat{j}\)
= (Δx)\(\hat{i}\) + (Δy)\(\hat{j}\)
Average velocity during that time interval is
\(\vec{v}_{av}=(\frac{Δx}{Δt})\hat{i}+(\frac{Δy}{Δt})\hat{j}=(v_{av})_x\hat{i}+(v_{av})_y\hat{j}\)
and is in the direction of the displacement \(Δ\vec{r}\)
Its magnitude is
\(|\vec{v}_{av}|=\sqrt{(v_{av})_x^2+(v_{av})_y^2}\)
and makes an angle θ with the +x-axis given by
tan θ = \(\frac{(v_{av})_y}{(v_{av})_x}=\frac{Δy/Δt}{Δx/Δt}\) = \(\frac{Δy}{Δx}\)
which is the slope of the secant joining the initial and final positions.
Instantaneous velocity,
\(\vec{v}=\lim_{Δt→0}\frac{Δ\vec{r}}{Δt}=\frac{d\vec{r}}{dt}\)
= \((\frac{dx}{dt})\hat{i}+(\frac{dy}{dt})\hat{j}\)
Its magnitude is
v = \(|\vec{v}|=\sqrt{(\frac{dx}{dt})^2+(\frac{dy}{dt})^2}\)
and makes an angle θ with the +x-axis given by
tan θ =\(\frac{dy/dt}{dx/dt}=\frac{dy}{dx}\)
which is the slope of the tangent to the curve at the point at which the instantaneous velocity is determined.
Average acceleration and instantaneous acceleration for the two-dimensional nonuniform motion of a particle :
Consider the two-dimensional nonuniform motion of a particle in the x-y plane. Its velocity-time graph is a curve, see fig.
Average acceleration,
If \(\vec{v}_1=v_{1x}\hat{i}+y_{1y}\hat{j}\) and \(\vec{v}_2=v_{2x}\hat{i}+y_{2y}\hat{j}\) are the instantaneous velocities at the instants t1 and t2, then the average acceleration during this time interval Δt = t2 — t1 is
\(\vec{a}_{av}=\frac{Δ\vec{v}}{Δt}=(\frac{Δv_x}{Δt})\hat{i}+(\frac{Δv_y}{Δt})\hat{j}=(v_{av})_x\hat{i}+(v_{av})_y\hat{j}\)
where, Δvx = v2x − v1x and Δvy = v2y − v1y with (aav)x and (aav)y being the x— and y- components of the average acceleration. The magnitude of the average acceleration is
\(|\vec{a}_{av}|=\sqrt{(a_{av})_x^2+(a_{av})_y^2}\)
and makes an angle θ with the +x-axis given by
tan θ = \(\frac{(a_{av})_y}{(a_{av})_x}= \frac{Δv_y}{Δv_x}\)
which is the slope of the secant joining the two points on the v-t graph.
Instantaneous acceleration,
\(\vec{a}=\lim_{Δt→0}\frac{Δ\vec{v}}{Δt}=\frac{d\vec{v}}{dt}\)
= \((\frac{dv_x}{dt})\hat{i}+(\frac{dv_y}{dt})\hat{j}\)
= \(\frac{d}{dt}(\frac{dx}{dt})\hat{i}+\frac{d}{dt}(\frac{dy}{dt})\hat{j}\)
= \((\frac{d^2x}{dt^2})\hat{i}+(\frac{d^2y}{dt^2})\hat{j}\)
= \(a_x\hat{i}+a_y\hat{j}\)
Its magnitude is a = |\(\vec{a}\)| = \(\sqrt{(a_x)^2+(a_y)^2}\)
and makes an angle θ with the +x-axis given by
tan θ = \(\frac{dv_y/dt}{dv_x/dt}= \frac{dv_y}{dv_x}\)
which is the slope of the tangent to the v-t curve at the point at which the instantaneous velocity is determined.
| Remember:
Motion in two dimensions can be resolved into two independent motions in mutually perpendicular directions. |
Relative velocity :
Motion is relative, i.e., motion of a body can be defined only with respect to an observer.
The velocity determined by a particular observer is called the velocity relative to that observer, or simply, relative velocity.
If \(\vec{v}_A\) and \(\vec{v}_B\) are the velocities of two bodies A and B, respectively, relative to the ground, then the velocity of A relative to B or as observed from B (i.e., by an observer attached to B) is
\(\vec{v}_{AB}=\vec{v}_A-\vec{v}_B\)
and the velocity of B relative to A is
\(\vec{v}_{BA}=\vec{v}_B-\vec{v}_A\)
= \(-\vec{v}_{AB}\)
Q. If , and are respectively the velocity of A relative to B, the velocity of B relative to C, and the velocity of C relative to D, what is the velocity of A relative to D?
Ans. The velocity of A relative to D, by definition,
\(\vec{v}_{AD}=\vec{v}_A-\vec{v}_D\)
∴ \(\vec{v}_{AD}=\vec{v}_A-\vec{v}_B+\vec{v}_B-\vec{v}_C+\vec{v}_C-\vec{v}_D\)
= \(\vec{v}_{AB}+\vec{v}_{BC}+\vec{v}_{CD}\)
Projectile Motion :
An object projected or launched with an initial velocity to move freely under the action of gravity in a vertical plane is called a projectile.
Examples :
- A stone thrown towards a hanging fruit.
- A basketball lobbed towards the basket.
- A football kicked by a player.
- A bullet or shell fired from a gun.
- A relief packet or bomb released from an aircraft in flight.
Range of a projectile : The range or horizontal range of a projectile is the horizontal distance it travels from the point of projection to the point on the same horizontal plane at the end of its trajectory.
OR
The range of a projectile is the horizontal distance it has travelled when it returns to its, initial launch height, usually the ground level.
Velocity of projection or launch velocity : The initial velocity with which a projectile is launched is called the velocity of projection or launch velocity.
Angle of projection or launch angle : The angle with the horizontal at which an object is projected or launched is called the angle of projection or launch angle. OR
The angle made by the velocity of projection or launch velocity with the horizontal is called the angle of projection.
Trajectory of a projectile : The curved path followed by a projectile in its two-dimensional motion is called the trajectory.
Expressions for Trajectory of a projectile :
Suppose a projectile is launched obliquely with an initial velocity \(\vec{u}\), inclined at an angle θ to the horizontal, called the launch angle or the angle of projection, and that after launching no force, except that due to gravity, acts on it.
Then, the projectile's trajectory or path is confined to a vertical plane containing \(\vec{u}\). Its acceleration, due only to gravity, is constant near the Earth's surface and always vertically downward. In the horizontal direction it is not acted upon by any force. Hence, along that direction, it has no acceleration, i.e., it moves with a constant velocity. We orient the x-axis horizontal and the y-axis vertical such that \(\vec{u}\) lies in the xy plane and take the launch point as the origin of the axes; the motion then lies in the xy plane. (For reference, see Fig.)
The x and y components of the kinematic variables are
(i) the initial velocity :
ux = u cos θ, uy = u sin θ …… (1)
(ii) the acceleration :
ax = 0, ay = −g ….(2)
(iii) the displacement :
x = uxt + ½ axt2 = (u cos θ)t …….(3)
y = uyt + ½ ayt2
= (u sin θ)t − ½ gt2 ……(4)
(iv) the instantaneous velocity at time t
vx = ux + axt = ux = u cos θ …….(5)
vy = uy + ayt = u sin θ — gt ………(6)
(v) the time of ascent :
The time of ascent (ta) is the time taken by the projectile to reach the peak of its trajectory where vy = 0. Hence, substituting t = ta and vy = 0 in the above equation (6),
ta = \(\frac{ u\,sin\,θ}{g}\)
(vi) the time of flight :
The time of flight (T) of a projectile is the time it takes to cover its entire trajectory or to cover its horizontal range
The time of descent (td) is the time taken by the projectile to return to its initial launch height, usually the ground level, from the peak of its trajectory. It can be shown that td = ta.
∴ T = ta + td = 2ta = \(\frac{2u\,sin\,θ}{g}\)
(vii) the horizontal range :
The range or horizontal range R is the horizontal distance travelled by the projectile during its flight.
∴ R = uxT = \(\frac{2u_xu_y}{g}=\frac{2(u\,sin\,θ\,cos\,θ)}{g}\)
= \(\frac{u^2(2\,sin\,θ\,cos\,θ)}{g}\)
∴ R = \(\frac{u^2(sin\,2θ)}{g}\)….(7)
(viii) the peak (i.e., maximum) height.
The vertical component of the displacement of the projectile at an instant t is
y = uyt − ½ gt2
At t= ta, the projectile is at its peak, i.e., at maximum height, y = H.
H = uyta − ½ gt2a
= \(u_y(\frac{u_y}{g})-\frac{1}{2}g(\frac{u_y}{g})^2\)
= \(\frac{u_y^2}{2g}=\frac{u^2sin^2θ}{2g}\) …….(8)
The path of a projectile, ignoring the effects of air, is a parabola :
Explanation :
From above Eq. (3), t = \(\frac{x}{u\,cos\,θ}\)
Substituting the value of t in Eq. (4),
y = (u sin θ)\(\frac{x}{u\,cos\,θ}-\frac{1}{2}g(\frac{x}{u\,cos\,θ})\)
∴ y = x tan θ\(-(\frac{g}{2u^2\,cos^2θ})x^2\)
Since θ, u and g are the constants of the motion, the above equation is of the form y = bx — cx2, in which b and c are constants. This being the equation of a parabola, the trajectory of the projectile is a parabola.
| Know this :
The expressions for T, R, Rmax and H are valid only if no force other than that due to gravity acts on the projectile. In reality, air resistance greatly affects these quantities, e.g., ta > td and Rmax < u2/g. In practice, therefore, the launch angle for maximum range should be greater than 45°. |
Q. Show that for a given launch velocity, the range of a projectile is the same for two complementary launch angles.
Ans. For a launch velocity and launch angle θ, the range of a projectile is,
R = \(\frac{u^2(sin\,2θ)}{g}=\frac{2u^2(sin\,θ\,cos\,θ)}{g}\)
Since sin (90°— θ) = cos θ and cos (90°— θ) = sin θ,
sin (90°— θ) cos (90°— θ) = cos θ sin θ = sin θ cos θ that is, the product sin θ cos θ remains the same when θ is replaced by (90°— θ).
Hence, for a given launch velocity of magnitude u, the projectile has the same range for two complementary launch angles:
Uniform circular motion :
Circular motion : The motion of a particle along a complete circle or a part of it is called circular motion.
Radius vector : For a particle performing circular motion, its position vector with respect to the centre of the circle is called the radius vector.
- The radius vector has a constant magnitude, equal to the radius of the circle. However, its direction changes as the position of the particle changes along the circumference
Angular displacement : The change in the angular position of a particle performing circular motion with respect to a reference line in the plane of motion of the particle and passing through the centre of the circle is called the angular displacement.
As the particle moves in its circular path, its angular position changes, say from θ1 at time t to θ2 at a short time δt later, see fig.
Angular displacement in circular motion
In the interval δt, the position vector sweeps out an angle δθ = θ2 — θ1. δθ is the magnitude of the change in the angular position of the particle.
Infinitesimal angular displacement in an infinitesimal time interval δt —>0, is given a direction perpendicular to the plane of revolution by the right hand rule or the right-handed screw rule (see below fig-5.).
Angular velocity : The time rate of angular displacement of a particle performing circular motion is called the angular velocity.
If the particle has an angular displacement δθ in a short time interval δt its angular velocity
\(\vec{ω} =\lim_{δt→0}\frac{\vec{δθ}}{dt}=\frac{\vec{dθ}}{dt}\)
is a vector in the direction of \(\vec{δθ}\), given by the right hand rule or right-handed screw rule.
Right hand rule and right-handed screw rule.
Period of revolution : The time taken by a particle performing UCM to complete one revolution is called the period of revolution or the period (T) of UCM.
Angular Speed : For a particle performing circular motion, the time rate at which the angle is swept out by the radius vector is called the angular speed.
In a uniform circular motion (UCM) of radius r, the particle covers a distance equal to the circumference of the circle in one period T and its angular position changes by 2p rad.
∴ Period, T =\(\frac{Circumference}{\text{linear speed}}=\frac{2πr}{v}\)
∴ Speed, v =\(\frac{2πr}{T}\)
Angular speed, ω = \(\frac{\text{Change in angular position in 1 rev}}{\text{Period of revolution}}\)
∴ ω = \(\frac{2π}{T}\)=\(\frac{2πr/T}{r}=\frac{v}{r}\)
∴ v = ω r, This is the relation between the linear and angular speeds of the particle.
UCM is an accelerated motion : A particle in uniform circular motion (UCM) moves in a circle or circular arc at constant linear speed v. The instantaneous linear velocity \(\vec{v}\) of the particle is along the tangential to the path, in the sense of motion of the particle. Since changes in direction, without change in its magnitude, there must be an acceleration that must be
(i) perpendicular to \(\vec{v}\)
(ii) constant in magnitude
(iii) at every instant directed radially inward, i.e., towards the centre of the circular path.
Hence, a UCM is an accelerated motion. Such a radially inward acceleration is called a centripetal (meaning, centre-seeking) acceleration.
Expression for Centripetal Acceleration:
Consider a particle performing uniform circular motion with a constant angular velocity along a circle in the xy plane. Let r be the radius of the circle and its centre be the origin of the coordinate axes. If the angular position of the particle at any instant t is θ as shown in the figure, then θ = ωt.
If x and y are the coordinates of the particle, its position vector or the radius vector at time t is \(\vec{r}=x\hat{i}+y\hat{j}\)
where \(\hat{i}\) and \(\hat{j}\) are unit vectors along the x and y axes respectively.
From Δ POM,
x =r cos θ = r cos ωt and
y = r sin θ = r sin ωt
∴ \(\vec{r}\) = r cos ωt \(\hat{i}\) + r cos ωt \(\hat{j}\)
The linear velocity of the particle is
\(\vec{v}=\frac{d\vec{r}}{dt}=\frac{d}{dt}\)(r cos ωt \(\hat{i}\) + r cos ωt \(\hat{j}\))
∴\(\vec{r}\) = −ωr sin ωt \(\hat{i}\) + ωr cos ωt \(\hat{j}\)
Since r and ω are constant in uniform circular motion is along the tangent to the path in the sense of motion of the particle.
The linear acceleration of the particle is
\(\vec{a}=\frac{d\vec{v}}{dt}=\frac{d}{dt}\)(−ωr sin ωt \(\hat{i}\) + ωr cos ωt \(\hat{j}\))
= −ω2r cos ωt \(\hat{i}\) − ω2r sin ωt \(\hat{j}\)
= −ω2(r cos ωt \(\hat{i}\) + r sin ωt \(\hat{j}\))
∴\(\vec{a}\)=−ω2\(\vec{r}\)
The acceleration of a particle performing UCM has the magnitude ω2r. The minus sign shows that its direction is opposite to that of \(\vec{r}\).
It means that at every instant the acceleration is directed along the radius towards the centre of the circle. This acceleration is called radial acceleration or centripetal acceleration.
A net real force must act on the particle to produce this acceleration. This force, which at every instant must point towards the centre of the circle, is called the centripetal force.
If m is the mass of the particle, the centripetal force,
\(\vec{F}\)= m\(\vec{a}\) = −mω2\(\vec{r}\)
In magnitude,
F = mω2r = mωv = mv2/r
Centripetal force : In the uniform circular motion of a particle, the centripetal force is the force on the particle which at every instant points radially towards the centre of the circle and produces the centripetal acceleration to move the particle in its circular path.
Conical pendulum : A conical pendulum is a simple pendulum whose bob revolves in a horizontal circle with constant speed with the centre of its circular path below the point of suspension such that the string makes a constant angle θ with the vertical.
Time period of conical pendulum :
Consider a conical pendulum of string length L with its bob of mass m performing UCM along a horizontal circle of radius r (Fig-7.).
The centre O’ of the circular path is vertically below the point of suspension O, and the string makes a constant angle θ with the vertical.
At every instant of its motion, the bob is acted upon by its weight m\(\vec{g}\) and the tension\(\vec{F}\) in the string. If the constant linear speed of the bob is v, the necessary horizontal centripetal force is
FC = mv2/r
\(\vec{F}\) is the resultant of the tension in the string and the weight. Resolve \(\vec{F}\) into components F cos θ vertically opposite to the weight of the bob and F sin θ horizontal. F cos θ balances the weight. F sin θ is the necessary centripetal force.
∴ F sin θ = mv2/r and
F cos θ = mg
Dividing Eq. (1) by Eq. (2),
tan θ = v2/rg ……(3)
The period T of the conical pendulum is the time taken by its bob to complete one revolution in a horizontal circle with constant speed v.
Then,
v = 2πr/T ….(4)
∴ tan θ = \(\frac{4π^2r^2}{T^2rg}\)=\(\frac{4π^2r}{T^2g}\)
∴ T2 = \(\frac{4π^2r}{g\,tan\,θ}\)= \(\frac{4π^2L\,sin\,θ}{g\,tan\,θ}\) ..(‘.’ r = L sin θ)
= \(\frac{4π^2L\,cos\,θ}{g}\)
∴ T = 2π\(\sqrt{\frac{L\,cos\,θ}{g}}\)=2π\(\sqrt{\frac{h}{g}}\) … (5)
where h = L cos θ is the axial height of the cone.
From the above expression, we can see that period of a conical pendulum depends on the various factors.
(i) T ∝ \(\sqrt{L}\)
(ii) T ∝ (\sqrt{cos\,θ}\) ….(If θ increases, cos θ and T decrease)
(iii) T ∝\(\frac{1}{\sqrt{g}}\)
(iv) The period is independent of the mass of the bob.
Linear speed of the bob of a conical pendulum :
From eq. (3)
v2 = rg tan θ
∴ v = \(\sqrt{rg\,tan\,θ}\) …. (6)
Expression for the angular speed of the bob :
From eq. (5)
T = 2π\(\sqrt{\frac{L\,cos\,θ}{g}}\)
If the angular speed of the bob is ω,
ω =2π/T = \(\sqrt{\frac{g}{L\,cos\,θ}}\)
This is the required expression. From the above equation,
cos θ = g/ω2L
Therefore, as ω increases, cos θ decreases and θ increases.
We reply to valid query.