I.E. Irodov Solutions on Dynamics OF A Solid Body

1. A thin uniform rod AB of mass m = 1.0 kg moves translationally with acceleration w = 2.0 m/s² due to two antiparallel forces F₁ and F₂ (Fig. 1.52). The distance between the points at which these forces are applied is equal to a = 20 cm. Besides, it is known that F₂ = 5.0 N. Find the length of the rod.

Since, motion of the rod is purely translational, net torque about the C.M of the rod should be equal to zero.

Thus F₁l/2 = F₂(l/2 – a)

Or F₁/F₂ 1 – a/(l/2) (1)

For the translational motion of rod

F₂ – F₁ = mw_c

Or 1 – F₁/F₂ = mw_c/F₂ (2)

From (1) and (2)

a/(l/2) = mw_c/F₂

Or l = 2aF₂/mw_c = 1m

2. A force F = Ai + Bj is applied to a point whose radius vector relative to the origin of coordinates O is equal to r = ai + bj, where a, b, A, B are constants, and i, j are the unit vectors of the x and y axes. Find the moment N and the arm l of the force F relative to the point O.

Sought moment

=

=

=

And arm of the force l = N/F = (aB – Ab)/√(A² + B²)

3. A force F₁ = Aj is applied to a point whose radius vector r_i = ai, while a force F₂ = Bi is applied to the point whose radius vector r₂ = bj . Both radius vectors are determined relative to the origin of coordinates O, i, and j are the unit vectors of the x and y axes, a, b, A, B are constants. Find the arm l of the resultant force relative to the point O.

Relative to point O, the net moment of force

=

=

Resultant of the external force

As

as

l = N/F = (ab – AB)/√(A² + B²)

4. Three forces are applied to a square plate as shown in Fig. 1.53. Find the modulus, direction, and the point of application of the resultant force, if this point is taken on the side BC.

For coplanar forces, about any point in the same plane

Where

Or

Thus length of the arm, l = N_net/F_net

Here obviously

Thus

Hence l = F(a/√2)/2F = a/2√2 = (a/2)sin 45⁰

Thus the point of application of force is at the midpoint of the side BC.

5. Find the moment of inertia

(a) of a thin uniform rod relative to the axis which is perpendicular to the rod and passes through its end, if the mass of the rod is m and its length l;

(b) of a thin uniform rectangular plate relative to the axis passing perpendicular to the plane of the plate through one of its vertices, if the sides of the plate are equal to a and b, and its mass is m.

(a)Consider a strip of length dx at a perpendicular distance x from the axis about which we have to find the moment of inertia of the rod. The elemental mass of the rod equals

dm = (m/l)dx

Moment of inertia of this element about the axis

dI = dmx² = (m/l)dx. x²

Thus, the moment of inertia of the rod, as a whole about the given axis

I = ∫₀^l (m/l)x² dx

= ml²/3

(b) Let us imagine the plane of the plate as xy plane taking the origin at the intersection point of the sides of the plate (Fig.).

Obviously I_x = ∫ dm y²

=

= ma²/3

Similarly I_y = mb²/3

Hence from the perpendicular axis theorem

I_z= I_x + I_y

= (m/3)(a² + b²), which is the sought moment of inertia.

8. A uniform disc of radius R = 20 cm has a round cut as shown in Fig. 1.54. The mass of the remaining (shaded) portion of the disc equals m = 7.3 kg. Find the moment of inertia of such a disc relative to the axis passing through its centre of inertia and perpendicular to the plane of the disc.

For simplicity let us use a mathematical trick. We consider the portion of the given disc as the superposition of two complete discs (without holes), one of positive density and radius R and the other of negative density but of the same magnitude and radius R/2.

As (area) α (mass), the respective masses of the considered discs are (4m/3) and (-m/3) respectively, and these masses can be imagined to be situated at their respective centers (C.M). Let us take point O as origin and point x-axis towards right. Obviously, the C.M. of the shaded position of the given shape lies on the x-axis. Hence the C.M. (C) of the shaded portion is given by

x_c = ((-m/3)(-R/2( + (4m/3)0)/((-m/3) + 4m/3)

= R/6

Thus C.M. of the shape is at a distance R/6 from point O toward x-axis.

Using parallel axis theorem and bearing in mind that the moment of inertia of a complete homogeneous disc of radius m₀ and radius r₀ equals ½ m₀r₀². The moment of inertia of the small disc of mass (-m/3) and radius R/2 about the axis passing through point C and perpendicular to the plane of the disc

I_2C = ½ (-m/3)(R/2)² + (-m/3)(R/2 + R/6)²

= -mR²/24 – (4/27)mR²

Similarly I_1C = ½ (4m/3)R² + (4m/3)(R/6)²

= ⅔ mR² + mR²/27

Thus the sought moment of inertia,

I_C = I_1C + I_2C = (15/24)mR² – (3/27)mR² = (37/72)mR²

Moment of inertia of the shaded portion, about the axis passing through its centre,

I = (⅖)(4/3)πR³ρR² – ⅖ (4/3) πr³ρr³

= ⅖ (4/3)πρ(R⁵ – r⁵)

Now, if R = r + dr, the shaded portion becomes a shell, which is the required shape to calculate the moment of inertia.

Now, I = ⅖ – (4/3) πρ( r + dr)⁵ – r⁵)

= ⅖ (4/3)πρ(r⁵ + 5r⁴dr + ….-r⁵)

Neglecting higher terms

= ⅔ (4πr² dr ρ)r²

= ⅔ mr²

10. A light thread with a body of mass m tied to its end is wound on a uniform solid cylinder of mass M and radius R (Fig. 1.55). At a moment t = 0 the system is set in motion. Assuming the friction in the axle of the cylinder to be negligible, find the time dependence of

(a) the angular velocity of the cylinder;

(b) the kinetic energy of the whole system.

(a) Net force which is effective on the system (cylinder M + body m) is the weight of the body m in a uniform gravitational field, which is a constant. Thus the initial acceleration of the body m is also constant.

From the conservation of mechanical energy of the said system in the uniform field of gravity at time t = ∆t: ∆T + ∆U = 0

Or ½ mv² + ½ (MR²/2)ω² – mg∆h = 0

Or ¼ (2m + M)v² – mg∆h = 0 [as v = ωR at all times] (1)

But v² = 2w∆h

Hence using it in Eq. (1), we get

¼ (2m + M)2w∆h – mg ∆h = 0 or w = 2mg/(2m + M)

From the kinematical relationship, β = w/R = 2mg/(2m + M)R

Thus the sought angular velocity of the cylinder ω(t) = βt = 2mgt/(2m + M)R = gt/(1 + M/2m)R

(b) sought kinetic energy

T(t) = ½ mv² + ½ (MI²/2) ω²

= ¼ (2m + M)R²ω²

11. The ends of thin threads tightly wound on the axle of radius r of the Maxwell disc are attached to a horizontal bar. When the disc unwinds, the bar is raised to keep the disc at the same height. The mass of the disc with the axle is equal to m, the moment of inertia of the arrangement relative to its axis is I. Find the tension of each thread and the acceleration of the bar.

For equilibrium of the disc and axle

2T = mg or T = mg/2

As the disc unwinds, it has an angular acceleration β given by

Iβ = 2Tr or β = 2Tr/I = mgr/I

The corresponding linear acceleration is

rβ = w = mgr²/I

Since the disc remains stationary under the combined action of this acceleration and the acceleration (-w) of the bar which is transmitted to the axle, we must have

w = mgr²/I

13. In the arrangement shown in Fig. 1.56 the mass of the uniform solid cylinder of radius R is equal to m and the masses of two bodies are equal to m₁ and m₂. The thread slipping and the friction in the axle of the cylinder are supposed to be absent. Find the angular acceleration of the cylinder and the ratio of tensions T₁/T₂ of the vertical sections of the thread in the process of motion.

First of all, let us sketch free body diagram of each body. Since the cylinder is rotating and massive, the tension will be different in both the sections of threads. From Newton’s law in projection form for the bodies m₁ and m₂ and noting that w₁ = w₂ = w = βR, (as no thread slipping), we have (m₁>m₂).

m₁g – T₁ = m₁w = m₁βR

And T₂ – m₂g = m₂w (1)

Now from the equation of rotational dynamics of a solid about stationary axis of rotation,

i.e. N_z= Iβ_z , for the cylinder.

Or (T₁ – T₂)R = Iβ = mR²β/2 (2)

Simultaneous solution of the above equation yields:

β = (m₁ – m₂)g/R(m₁ + m₂ + m/2) and T₁/T₂ = m₁(m+4m₂)/m₂(m+4m₁)

14. In the system shown in Fig. 1.57 the masses of the bodies are known to be m₁ and m₂ , the coefficient of friction between the body m₁ and the horizontal plane is equal to k, and a pulley of mass m is assumed to be a uniform disc. The thread does not slip over the pulley. At the moment t = 0 the body m₂ starts descending. Assuming the mass of the thread and the friction in the axle of the pulley to be negligible, find the work performed by the friction forces acting on the body m₁ over the first t seconds after the beginning of motion.

As the system (m + m₁ + m₂) is under constant forces,the acceleration of body m₁ and m₂ is constant. In addition to it the velocities and accelerations of bodies m₁ and m₂ are equal in magnitude (say v and w) because the length of the thread is constant. From the equation of increment of mechanical energy i.e. ∆T + ∆U = A_fr, at time t when block m₁ is distance h below from initial position corresponding to t = 0.

½ (m₁ + m₂)v² + ½ (mR²/2)v²/R² – m₂gh = -km₁gh (1)

(as angular velocity ω = v/R for no slipping of thread.)

But v² = 2wh

So using it in (1), we get

w = 2(m₂ – km₁)g/(m + 2(m₁ + m₂)) (2)

Thus the work done by the friction force on m₁

A_fr = -km₁gh = -km₁g(½ wt²)

= -km₁(m₁– km₁)g²t²/(m + 2(m₁ + m₂)) (using 2)

15. A uniform cylinder of radius R is spinned about its axis to the angular velocity ω₀ and then placed into a corner (Fig. 1.58). The coefficient of friction between the corner walls and the cylinder is equal to k. How many turns will the cylinder accomplish before it stops?

In the problem, the rigid body is in translation equilibrium but there is an angular retardation. We first sketch the free body diagram of the cylinder.

Obviously the friction forces, acting on the cylinder, are kinetic. From the condition of translational equilibrium for the cylinder,

mg = N₁ + kN₂;

N₂ = kN₁

Hence N₁ = mg/(1 + k²)

N₂ = kmg/(1 + k²)

For pure rotation of the cylinder about its rotation axis,

N_z = Iβ_z

Or, -kN₁R – kN₂R = mR²β_z/2

Or, -kmgR(1+k)/(1+k²) = mR²β_z/2

Or, β_z = -2k(1+k)g/(1+k²)R

Now, from the kinematical equation,

ω² = ω₀² + 2β_z∆ϕ we have

∆ϕ = ω₀²(1+k²)R/4k(1+k)g, because ω = 0

Hence the sought number of turns,

n = ∆ϕ/2π = ω₀²(1+k²)R/8πk(1+k)g

18. A uniform cylinder of radius R and mass M can rotate freely about a stationary horizontal axis O (Fig. 1.59). A thin cord of length l and mass m is wound on the cylinder in a single layer. Find the angular acceleration of the cylinder as a function of the length x of the hanging part of the cord. The wound part of the cord is supposed to have its centre of gravity on the cylinder axis.

Let us use the equation dM_z/dt = N_zrelative to the axis through O (1)

For this purpose, let us find the angular momentum of the system M_z about the given rotation axis and the corresponding torque N_z. The angular momentum is

M_z = Iω + mvR

= (m₀/2 + m)R²ω

(where I = m₀R²/2 and v = ωR (no cord slipping)

So dM_z/dt = (MR²/2 + mR²)β_z (2)

The downward pull of gravity on the overhanging part is the only external force, which exerts a torque about the z axis, passing through O and is given by,

N_z =(m/l)xgR

Hence from the equation dM_z/dt = N_z

(MR²/2 + mR²)β_z= (m/l)xgR

Thus, β_z= 2mgx/lR(M+2m) > 0

Note: We may solve this problem using conservation of mechanical energy of the system (cylinder + thread) in the uniform field of gravity.

19. A uniform sphere of mass m and radius R rolls without slipping down an inclined plane set at an angle α to the horizontal.

Find:

(a) the magnitudes of the friction coefficient at which slipping is absent;

(b) the kinetic energy of the sphere t seconds after the beginning of motion.

(a) Let us indicate the forces acting on the sphere and their points of application. Choose positive direction of x and ϕ (rotation angle) along the incline in downward direction and in the sense of

i.e. F_x = mw_cx and N_cz = I_cβ_z we get:

mg sin α – f_r = mw (1)

And frR = ⅖ mR²β (2)

But fr ≤ kmg cos α (3)

In addition, the absence of slipping provides the kinematical relationship between the accelerations : w = βR (4)

The simultaneous solution of all the four equations yields:

k cos α ≥ 2/7 sin α

Or k ≥ 2/7 tan α

(b) solving Eqs. (1) and (2) [of part (a)], we get:

w_c = (5/7)g sin α

As the sphere starts at t = 0 along positive x axis, for pure rolling

v_c(t) = w_ct = (5/7)g sin αt (5)

Hence the sought kinetic energy

T = ½ mv_c² + ½ (⅖)mR²ω² = (7/10)mv_c² (as ω = v_c/R)

= (7/10)m(5g sin αt/7)²

= (5/14)mg² sin²α t²