Derivations

Derivations

Chapter 1. Rotational dynamics

Q. 1. Derive an expression for the maximum safe speed for a vehicle on a curved horizontal road. State its significance. (3 marks)

Ans. Consider a car of mass m taking a turn of radius along a horizontal (level) road. If µ, is the coefficient of static friction between the car tyres and the road surface, the limiting force of friction is µmg.

Then, the maximum safe speed max with which the car can take the turn without skidding off is determined by the condition,

maximum centripetal force = limiting force of static friction Mol

mo

r

= mg

max=Hs8

VS

Umax=rlsg

This is the required expression.

Significance: The above expression shows that the maximum safe speed depends critically upon friction which changes with the nature of the surfaces and presence of oil or water on the road. If the friction is not sufficient to provide the necessary centripetal force, the vehicle is likely to skid off the road.

Q. 2. A road at a bend should be banked for an optimum or most safe speed vo. Derive an expression for the required angle of banking. OR Obtain an expression for the optimum or most safe speed with which a vehicle can be driven along a curved banked road. Hence show that the angle of banking is independent of the mass of a vehicle. (4 marks)

Ans. Consider a car taking a left turn along a road of radius r banked at an angle for a designed optimum or most safe speed \mathcal{D}_{n} Let m be the mass of the car. In general, the forces acting on the car are

(a) its weight mg, acting vertically down

(b) the normal reaction of the road \vec{N}. perpendicular to the road surface c) the frictional force \vec{f}_{s} ( along the inclined surface of the road.

At the optimum speed, frictional force is not relied upon to contribute to the necessary lateral centripetal force. So, ignoring \vec{f}_{s}, resolve \vec{N} into two perpendicular components: Neos vertically up and Nsin horizontally towards the centre of the circular path. Since there is no acceleration in the vertical direction, Neos @ balances mg and Nsin provides the necessary centripetal force.

Resolve \vec{N} and \vec{f}_{S} into two perpendicular components: Ncos and f sin vertically up; Nsin & horizontally towards the centre of the circular path while f, cos horizontally outward. So long as the car takes the turn without sliding down, the sum Ncos 9+f sin balances mg, and N~sin~\theta-f_{s} cos 0 provides the necessary centripetal force. If v_{\mathfrak{mi}} is the minimum safe speed without skidding, mg=N~cos~\theta+f_{s}sin~\theta =N (cos 0+ µ, sin 0) and\frac{mv_{min}^{2}}{r}=N~sin~\theta-f_{s}cos~\theta =N(sin~\theta-\mu_{s}cos~\theta) ... Dividing Eq.(2) by Eq. (1), \frac{v_{min}^{2}}{rg}=\frac{sin~\theta-\mu_{s}cos~\theta}{cos~\theta+\mu_{s}sin~\theta}=\frac{tan~\theta-\mu_{s}}{1+\mu_{s}tan~\theta} v_{min}=\sqrt{\frac{rg(tan~\theta-\mu_{s})}{1+\mu_{s}tan~\theta}} Equation (3) gives the required expression for the minimum speed. 

the

(ii) For maximum safe speed: Refer to the answer to Solved Model Question Paper in Part 1. School

Q. 4. Derive an expression for the angular conical pendulum. of the bob of a

(3 marks) OR

Derive an expression for the frequency of a conical pendulum. of revolution of the bob of

5^{\circ}

(4 marks)

Ans. Consider a conical pendulum string length L with its bob of mass m performing UCM along a circular path of radius ras shown in the figure.

At every instant of its motion, the bob is acted upon by its weight m\vec{g} and the tension \vec{F} the string. If the constant angular speed of the bob is w, the necessary horizontal centripetal force is F_{c}=m\omega^{2}r

F is the resultant of the tension in the string and the weight. Resolve Finto components Fcos vertically opposite to the weight of the bob

[Note: From Eq. (4), cos theta = g / c * o ^ 2 * L Therefore, as increases, cos decreases and increases.]

If n is the frequency of revolution of the bob, omega = 2pi*n = sqrt(g/(L * cos theta)) n = 1/(2pi) * sqrt(g/(L * cos theta)) is the required expression for the frequency.

Q. 5. In the vertical circular motion of a body controlled by gravity, prove that the difference between the extreme tensions (or normal forces) depends only upon the weight of the body. (3 marks) OR A small body is tied to a string and revolved in a vertical circle of radius r. Prove that the difference in the tensions in the string at the highest and the lowest points is 6 times the weight of the body.

(3 marks)

Ans. Consider a small body (or particle) of mass m tied to a string and revolved in a vertical circle of radius r at a place where the acceleration due to gravity is g. At every instant of its motion, the body is acted upon by two forces, namely, its weight m vec g and the tension vec T in the string the figure. string a cho as shown in fig

Therefore, the difference in the tensions in the string at the highest and the lowest points depends only on the weight of the body and is equal to 6 times the weight of the body.

Q. 6. State and prove the theorem of parallel axis about moment of inertia.

(3 marks)

Ans. Theorem of parallel axis: The moment of inertia of a body about an axis is equal to the sum of (i) its moment of inertia about a parallel axis through its centre of mass and (ii) the product of the mass of the body and the square of the distance between the two axes.

Proof: Let I CM be the moment of inertia (MI) of a body of mass M about an axis through its centre of mass C, and I be its MI about a parallel axis through any point O. If the distance between the two axes is h, then the theorem of parallel axis can be stated mathematically as I=I CM +Mh^ 2

Consider an infinitesimal volume element of mass dm of the body at a point P. It is at a perpendicular distance CP from the rotation axis through C and a perpendicular distance OP from the parallel axis through O.

The MI of the element about the axis through C is C * P ^ 2 * dm Therefore, the MI of the body about the axis through the CM is I CM = int CP^ 2 dm. Similarly, the MI of the body about the parallel axis through O is I = integrate O * P ^ 2 dm

Draw PQ perpendicular to OC produced as shown in the figure. Then, from the figure, I = integrate O * P ^ 2 dm = integrate (O * Q ^ 2 + P * Q ^ 2) dm = mathfrak integrate [(OC + CQ) ^ 2 + P * Q ^ 2] dm = integrate (O * C ^ 2 + 2OCCQ + C * Q ^ 2 + P * Q ^ 2) dm = integrate (O * C ^ 2 + 2OCCQ + C * P ^ 2) dm (* = integrate O * C ^ 2 dm + integrate 2OCCQ dm + integrate C * P ^ 2 dm = O * C ^ 2 * integrate 1 dm + 2OC * integrate CQ dm + integrate C * P ^ 2 dm Since, OC = h is constant and integrate 1 dm = M is the mass of the vouy, I= M * h ^ 2 + 2h * integrate CQ dm +I CM Now, from the definition of centre of mass, the integral JCQ dm is mass M times a coordinate of the CM with respect to the origin C. Since C is itself the CM, this coordinate is zero and so also the integral. . I=I CM +Mh^ 2 C * Q ^ 2 + P * Q ^ 2 = C * P ^ 2 )

This proves the theorem of parallel axis.

Q. 7. State and prove the theorem of perpendicular axes.

(3 marks)

Ans. Theorem of perpendicular axes: The moment of inertia of a plane lamina about an axis perpendicular to its plane is equal to the sum of its moments of inertia about two mutually perpendicular axes in its plane and through the point of intersection of the perpendicular axis and the lamina. 

moments of inertia of the lamina about the x, y and z axes respectively, then, the theorem of perpendicular axes can be stated mathematically as I_{z} = I_{x} + I_{v} .

Consider an infinitesimal volume element of mass dm of the lamina at the point P(x,y) The MI of the lamina about the z-axis is I_{z} = integrate O * P ^ 2 dn . The element is at perpendicular distance y and x from the x- and y-axes respectively. Hence, the moments of inertia of the lamina about the x- and y-axes are respectively I_{x} = integrate y ^ 2 dm and I_{y} = integrate x ^ 2 dm . SinceOP^ 2 = y ^ 2 + x ^ 2 I_{x} = integrate O * P ^ 2 dm = integrate (y ^ 2 + x ^ 2) dm = integrate y ^ 2 dm + integrate x ^ 2 dm

therefore I_{z} = I_{x} + I_{1}

This proves the theorem of perpendicular axes.

Q. 8. State the expression for the moment of inertia of a thin uniform disc about an axis perpendicular to its plane and through its centre. Hence deduce the expression for its moment of inertia about a tangential axis perpendicular to its plane.

(2 marks)

Ans. (1) MI about the transverse symmetry axis: Consider a thin uniform disc of radius R and mass M. The axis of rotation through its centre C is perpendicular to its plane. C is also its centre of mass (CM). The MI of the disc about the transverse I CM = 1 2 MR^ 2 symmetry axis is ASC

(2) MI about a tangent perpendicular to its plane: Let / be its MI about an axis parallel to the transverse symmetry axis and tangent to the disc. Here, h = R = distance between the two axes. By the theorem of parallel axis, I=I CM +Mh^ 2 = (M * R ^ 2)/2 + M * R ^ 2 = 3/2 * M * R ^ 2

Q. 9. Assuming the expression for the moment of inertia of a ring about its transverse symmetry axis obtain the expression for its moment of inertia about (1) a diameter (2) a tangential axis in its plane.

(2 marks each)

Ans. Let M be the mass of a thin ring of radius R. Let I CM be the moment of inertia (MI) of the ring about its transverse symmetry axis. Then, I CM =MR^ 2

 MI about a diameter: Letx and y-axes be along two perpendicular diameters of the ring as shown in Fig. 1. Let I_{x} I_{v} and I_{z} be the moments of inertia of the ring about the x, y and z axes, respectively