Students can find the solutions of Problems In General Physics I.E. Irodov – The Fundamental Equation of Dynamics on this page. The Fundamental Equation of Dynamics is an important part of Physical Fundamentals of Mechanics. The branch of Physics which deals with the study of motion when subjected to forces or displacements and the subsequent effects of the bodies on their environment is called mechanics. The important topics of the Fundamental Equation of Dynamics are the fundamental equation of dynamics of a mass point (Newton’s second law), the same equation expressed in projections on the tangent and the normal of the point’s trajectory, the equation of dynamics of a point in the non-inertial reference frame, etc.
Learning these solutions will help students to score more marks in JEE exam and board exams. Students can improve their speed and accuracy by revising these solutions.
1. An aerostat of mass m starts coming down with a constant acceleration w. Determine the ballast mass to be dumped for the aerostat to reach the upward acceleration of the same magnitude. The air drag is to be neglected.
Let R be the constant upward thurst on the aerostat of mass m, coming down with a constant acceleration w. Applying Newton’s second law of motion for the aerostat in projection form
Fy = mwy
mg – R = mw …(1)
Now, if ∆m be the mass, to be dumped, then using the Eq. Fy = mwy
R – (m – ∆m)g = (m – ∆m)w ..(2)
From Eqn. (1) and (2), we get ∆m = 2mw/(g + w)
2. In the arrangement of Fig. the masses mo, m1, and m2 of bodies are equal, the masses of the pulley and the threads are negligible, and there is no friction in the pulley. Find the acceleration w with which the body mo comes down, and the tension of the thread binding together the bodies m1 and m2, if the coefficient of friction between these bodies and the horizontal surface is equal to k. Consider possible cases.
Let us write the fundamental equation of dynamics for all the three blocks in terms of projections, having taken the positive direction of x and y axes as shown in Fig; and using the fact that kinematical relation between the accelerations is such that the blocks move with same value of acceleration (say w)
m0g – T1 = m0w (1)
T1 – T2 – km1g = m1w (2)
And T2 – km2g = m2w (3)
The simultaneous solution of equations (1), (2) and (3) yields
w = g(m0 – k(m1 + m2))/(m0 + m1 + m2)
And T2 = (1 + k)m0 m2g/(m0 + m1 + m2)
As the block m0 moves down with acceleration w, so in vector form
3. Two touching bars 1 and 2 are placed on an inclined plane forming an angle α with the horizontal (Fig. 1.10). The masses of the bars are equal to m1 and m2, and the coefficients of friction between the inclined plane and these bars are equal to k1 and k2 respectively, with k1 > k2. Find:
(a) the force of interaction of the bars in the process of motion;
(b) the minimum value of the angle α at which the bars start sliding down.
Let us indicate the positive direction of x-axis along the incline (Fig.). Figures show the force diagram for the blocks. Let, R be the force of interaction between the bars and they are obviously sliding down with the same constant acceleration w
Newton’s second law of motion in projection form along x-axis for the blocks gives :
m1g sin α – k1m1 g cos α + R = m1w …(1)
m2g sin α – R – k2m2 g cos α = m2w …(2)
Solving Eqs. (1) and (2) simultaneously, we get
w = g sin α – g cos α(k1m1 + k2m2)/(m1 + m2) and
R = m1m2(k1– k2)g cos α/(m1 + m2) ..(3)
(b) when the blocks just slide down the plane, w = 0 , so from Eqn. (3)
g sin α – g cos α (k1m1 + k2m2)/(m1 + m2) = 0
Or (m1 + m2)sin α = (k1m1 + k2m2)cos α
Hence tan α = (k1m1 + k2m2)/(m1 + m2)
4. A small body was launched up an inclined plane set at an angle α = 150 against the horizontal. Find the coefficient of friction, if the time of the ascent of the body is η = 2.0 times less than the time of its descent.
Case 1. When the body is launched up :
Let k be the coefficient of friction, u the velocity of projection and l the distance traversed along the incline. Retarding force on the block = mg sin α + k mg cos α
and hence the retardation = g sin α + kg cos α .
Using the equation of particle kinematics along the incline,
0 = u2 -2 (g sin α + kg cos α) l
Case 2. When the block comes downward, the net force on the body = mg sin α – km g cos α and hence its acceleration = g sin α – kg cos α
Let, t be the time required then
l = ½ (g sin α – kg cos α)t’2 ..(4)
From Eqns (3) and (4)
t2/t’2 = (sin α – k cos α)/(sin α + k cos α)
But t/t’ = 1/η (according to question)
Hence on solving we get
k = (η2 – 1)/(η2 + 1) tan α = 0.16
5. The following parameters of the arrangement of Fig. 1.11 are available: the angle α which the inclined plane forms with the horizontal, and the coefficient of friction k between the body m1 and the inclined plane. The masses of the pulley and the threads, as well as the friction in the pulley, are negligible. Assuming both bodies to be motionless at the initial moment, find the mass ratio m2/m1 at which the body m2
(a) starts coming down;
(b) starts going up; (c) is at rest.
At the initial moment, obviously, the tension in the thread connecting m1 and m2 equals the weight of m2.
(a) For the block m2 to come down or the block m1 to go up, the conditions is
m2g – T ≥ 0 and T – m1g sin α – fr ≥ 0
where T is tension and fr is friction which in the limiting case equals km1g cosα. Then
m2g – m1 sin α > km1 g cos α
Or m2/m1 > (k cos α + sin α)
(b) Similarly in the case
m1g sin α – m2g > frlim
Or m1g sin α – m2g > km1g cos α
Or m2/m1 < sin α – k cos α
(c) For this case, neither kind of motion is possible, and fr need not be limiting.
Hence, (k cos α + sin α)> m2/m1 > (sin α – k cos α)
From the conditions, obtained in the previous problem, first, we will check whether the mass m2 goes up or down. Here, m2/ m1 = η > sin α + k cos α , (substituting the values). Hence the mass m2 will come down with an acceleration (say w). From the free body diagram of previous problem,
m2 – g -T = m2w (1)
And T – m1g sin α – km1g cos α = m1w (2)
Adding (1) and (2), we get
m2g – m1g sin α – km1g cos α = (m1 + m2)w
w = (m2/m1 – sin α – k cos α)g/(1 + m2/m1)
= (η – sin α – k cos α)g/(1+η)
Substituting all the values, w = 0.048 g ≈ 0.05 g
As m2 moves down with acceleration of magnitude w = 0.05 g > 0, thus in vector form acceleration of m2:
=
Let us write the Newton’s second law in projection form along the positive x-axis for the plank and the bar fr = m1w1 , fr = m2v2
At the initial moment, fr represents the static friction, and as the force F grows so does the friction force fr, but up to its limiting value.
I.e. fr = frs(max) = kN = km2g
Unless this value is reached, both bodies move as a single body with equal acceleration. But as soon as the force fr reaches the limit, the bar starts sliding over the plank i.e. w1 ≥ w2.
Substituting here the values of w1 and w2 taken from Eq. (1) and taking into account that
fr = km2g, we obtain, (at – km2g)/m2 ≥ km2g/m1, where the sign = corresponds to the moment t = t0 (say)
Hence t0 = kgm2 (m1 + m2)/am1
If t ≤ t0, then w1 = km2g/m2 (constant) and w2 = (at – km2g)/m2
On this basis w1 (r) and w2 (t), plots are as shown in the figure of answer sheet.
Let us designate the x-axis (Fig.) and apply Fx = mwx for body A:
mg sin α – kmg cos α = mw
Or w = g sin α – kg cos α
Now from kinematical equation:
l sec α = 0 + (½)wt2
Or t = √(2l sec α/(sin α – k cos α)g)
= √(2l/(sin 2α/2 – k cos2 α)g)
Using Eq. (1)
For tmin d(sin 2α/2 – k cos2α)/dα = 0
i.e. 2 cos 2α/2 +2k cos α sin α = 0
Or tan 2α = -1/k => α = 490
and putting the values of α , k and l in Eq . (2) we get tmin = 1s.
Let us fix the x-y coordinate system to the wedge, taking the x-axis up, along the incline, and the y-axis perpendicular to it (Fig.). Now, we draw the free body diagram for the bar.
Let us apply Newton’s second law in projection form along x and y axis for the bar :
T cos β – mg sin α – fr = 0 (1)
T sin β + N – mg cos α = 0
Or N = mg cos α – T sin β (2)
But fr = kN and using (2) in (1), we get
T = mg sin α + kmg cos α/(cos β + k sin β) (3)
For Tmin the value of (cos β + k sin β) should be maximum.
So, d(cos β + k sin β)/dβ = 0 or tan β = k
Putting this value of β in Eq.(3) we get,
Tmin = mg (sin α + k cos α)/1/√(1 + k2) + k2/√(1+k2)
= mg (sin α + k cos α)/√(1+k2)
10. At the moment t = 0 the force F = at is applied to a small body of mass m resting on a smooth horizontal plane (a is a constant). The permanent direction of this force forms an angle α with the horizontal (Fig. 1.14). Find:
(a) the velocity of the body at the moment of its breaking off the plane;
(b) the distance traversed by the body up to this moment.
First of all let us draw the free body diagram for the small body of mass m and indicate x-axis along the horizontal plane and y-axis, perpendicular to it, as shown in the figure. Let the block breaks off the plane at t = t0 i.e. N = 0
So N = mg – at0 sin α = 0
Or t0 = mg/a sin α (1)
From Fx = mwx, for the body under investigation:
mdvx/dt = at cos α; Integrating within the limits for v(t)
m ∫0vdvx = a cos α∫t dt (using (1))
So v = ds/dt = a cos α t2/2m (2)
Integrating Eqn (2) for s(t)
s = (a cos α/2m)(t3/3)
Using the value of t = t0 from Eq. (1) into Eqs. (2) and (3)
v = mg2cos α/2a sin2α
And s = m2g3cos α/6a2 sin3α
Newton’s second law of motion in projection form, along horizontal or x-axis i.e
Fx = mwx gives
F cos (as) = mv dv/ds (as α = as)
Or F cos (as)ds = mv dv
Integrating, over the limits for v(s)
Or v = √(2F sin α/ma)
= √(2g sin α/3a (using F = mg/3) which is the sought relationship.
From the Newton’s second law in projection from :
For the bar,
T – 2 kmg = (2 m) w ..(1)
For the motor,
T – kmg = m w’ .. (2)
Now, from the equation of kinematics in the frame of bar or motor
l = ½ (w + w’)t2 ..(3)
From (1), (2) and (3) we get on eliminating T and w’
t = √2l/(kg + 3w)
Let us write Newton’s second law for both, bar 1 and body 2 in terms of projection having taken the positive direction of x1 and x2 as shown in the figure and assuming that body 2 starts sliding, say, upward along the incline
T1 – m1g sin α = m1w1 (1)
m2g – T2 = m2w (2)
For the pulley, moving in vertical direction
from the equation Fx = mwx
2T2 – T1 = (mp)w1 = 0
(as mass of the pulley mp = 0 )
Or T1 = 2T2 ..(3)
As the length of the threads are constant, the kinematical relationship of accelerations becomes w = 2w1 ..(4)
Simultaneous solutions of all these equations yields:
w = 2g(2m2/m1 – sin α)/(4m2/m1 + 1)
= 2g(2η – sin α)/(4η + 1)
As η > 1, w is directed vertically downward, and hence in vector form
Let us write Newton’s second law for masses m1 and m2 and moving pully in vertical direction along positive x – axis (Fig.) :
m1g – T = m1w1x (1)
m2g – T = m2w2x (2)
T1 – 2T = 0 (as m = 0)
T1 = 2T (3)
Again using Newton’s second law in projection form for mass m0 along positive x1 direction (Fig.), we get
T1 = m0 w0 (4)
The kinematical relationship between the accelerations of masses gives in terms of projection on the x-axis
w1x + w2x = 2w0 (5)
Simultaneous solution of the obtained five equations yields:
w1 = [4m1m2 +m0(m1 – m2)]g/4m1m2 + m0(m1 + m2)
In vector form
As the thread is not tied with m, so if there were no friction between the thread and the ball m, the tension in the thread would be zero and as a result both bodies will have free fall motion. Obviously, in the given problem, it is the friction force exerted by the ball on the thread, which becomes the tension in the thread. From the condition or language of the problem wM > wm and as both are directed downward so, relative acceleration of M = wM – wm and is directed downward. Kinematical equation for the ball in the frame of rod in projection form along upward direction gives :
l = ½ (wM – wm)t2 (1)
Newton’s second law in projection form along vertically down direction for both, rod and ball gives,
Mg – fr = MwM (2)
mg – fr = mwm (3)
Multiplying Eq. (2) by m and Eq. (3) by M and then subtracting Eq. (3) from (2) and after using Eq. (1) we get
fr = 2lMm/(M – m)t2
Suppose, the ball goes up with acceleration w1 and the rod comes down with the acceleration w2.
As the length of the thread is constant, 2w1 = w2 ..(1)
From Newton’s second law in projection form along vertically upward for the ball and vertically downward for the rod respectively gives,
T – mg = mw1 (2)
And Mg – T’ = Mw2 (3)
but T = 2T (because pulley is massless) (4)
From Eqs. (1), (2), (3) and (4)
w1 = (2M – m)g/(m + 4M)
= (2- η)g/(η+4) (in upward direction)
And w2 = 2(2 – η)g/(η+4) (downwards)
From kinematical equation in projection form, we get
l = ½ (w1 + w2)t2
as, w1 and w2 are in the opposite direction. Putting the values of w1 and w2, the sought time becomes
t = √(2l(η +4)/3(2 – η)g) = 1.4s
Using Newton’s second law in projection form along x – axis for body 1 and along negative x-axis for body 2 respectively, we get
m1g – T1 = m1w1 (1)
T2 – m2g = m2w2 (2)
For the pulley lowering in the downward direction from Newton’s law along the x-axis,
T1 – 2T2 = 0 (as pulley is massless)
Or T1 = 2T2 (3)
As the length of the thread is constant so
w2 = 2w1 (4)
The simultaneous solution of above equations yields,
w2 = 2(m1 – 2m2)g/(4m2 + m1)
= 2(η -2)/(η + 4) (as m1/m2 = η)
Obviously, during the time interval in which the body 1 comes to the horizontal floor covering the distance h, body 2 moves upward the distance 2h. At the moment when body 2 is at the height 2h from the floor its velocity is given by the expression :
V22 = 2w2(2h) = 2(2(η-2)2gh/(η+4)
= 8h(η-2)g/(η+4)
After the body m1 touches the floor the thread becomes slack or the tension in the thread zero, thus as a result body 2 is only under gravity for it’s subsequent motion. Owing to the velocity v2 at that moment or at the height 2h from the floor, the body 2 further goes up under gravity by the distance,
h’ = v22/2g = 4h(η-2)/(η+4)
Thus the sought maximum height attained by the body 2 :
H = 2h + h’ = 2h + 4h(η-2)/(η+4)
= 6ηh/(η+4)
18. Find the accelerations of rod A and wedge B in the arrangement shown in Fig. 1.20 if the ratio of the mass of the wedge to that of the rod equals η, and the friction between all contact surfaces is negligible.
Let us draw free body diagram of each body, i.e. of rod A and of wedge B, and also draw the kinematical diagram for accelerations, after analyzing the directions of motion of A and B. Kinematical relationship of accelerations is :
tan α = wA/wB
Let us write Newton’s second law for both bodies in terms of projections having taken positive directions of y and x-axes as shown in the figure.
mAg – N cos α = mAwA (2)
And N sin α = mBwB (3)
Simultaneous solution of (1), (2) and (3) yields :
wA = mAg sin α/(mA sin α + mB cot α cos α) = g/(1 + η cot2 α) and
wB = wA/tan α = g/(tan α + η cot α)
Note : We may also solve this problem using conservation of mechanical energy instead of Newton’s second law.
Let us draw free body diagram of each body and fix the coordinate system, as shown in the figure. After analysing the motion of M and m on the basis of force diagrams, let us draw the kinematical diagram for accelerations (Fig.).
As the length of threads are constant so,
dsmM = dsM and as
So from the triangle law of vector addition
wm = √2w (1)
From the Eq. Fx = mwx , for the wedge and block :
T – N = Mw, (2)
and N = mw (3)
Now, from the Eq. Fy = mwy , for the block
mg – T – kN= mw (4)
Simultaneous solution of Eqs. (2), (3) and (4) yields :
w = mg/(km + 2m + M)
= g/(k + 2 + M/m)
Hence using Eq. 1
wm = g√2/(2 + k + M/m)
Bodies 1 and 2 will remain at rest with respect to bar A for wmin ≤ w ≤ wmax, where wmin is the sought minimum acceleration of the bar. Beyond these limits, there will be a relative motion between bar and the bodies. For 0 ≤ w≤ wmin, the tendency of body 1 in relation to the bar A is to move towards right and is in the opposite sense for w ≥ wmax. On the basis of the above argument the static friction on 2 by A is directed upward and on 1 by A is directed towards left for the purpose of calculating wmin.
Let us write Newton’s second law for bodies 1 and 2 in terms of projection along positive x-axis (Fig.).
T-fr1 = mw or fr1 = T – mw (1)
N2 = mw (2)
As body 2 has no acceleration in vertical direction, so
fr2 = mg – T (3)
From (1) and (3)
(fr1 + fr2) = m(g – w) (4)
But fr1 + fr2 ≤ k(N1 + N2)
Or fr1 + fr2 ≤ k(mg + mw) (5)
From (4) and (5)
m(g – w) ≤ mk(g + w)
Or w ≥ g(1-k)/(1+k)
wmin = g(1-k)/(1+k)
Comments