Mechanics
Exam Duration: 45 Mins Total Questions : 30
A particle slides down a smooth inclined plane of elevation 45° fixed in an elevator going up with an acceleration g. The base of incline has length L.The time taken by particle to reach the bottom is
- (a)
\(\sqrt{L\over g}\)
- (b)
\(\sqrt{2L\over g}\)
- (c)
\(\sqrt{L\over 2g}\)
- (d)
\(\sqrt{g\over L}\)
Draw FBD,
By Newton's law,
mg sin θ + maosinθ = ma
or a = (g + ao) sin θ
Given, ao = g and θ = 45° so, a =\(2g\over \sqrt{2}\)
As \(s=0+{1\over2}at^2\)
\(\therefore\ {L\over cos\ 45^0}={1\over2}\times{2g\over \sqrt2}\times t^2\)
\(t=\sqrt{2L\over g}\)
A roller of weight 500 N has a radius of 120 mm and is pulled over a step of height 60 mm by a horizontal force P. Find magnitude of P to just start the roller over the step. (Refer figure)
- (a)
210N
- (b)
279N
- (c)
288N
- (d)
291N
Applying Lami's theorem
\({P\over sin150^0}={500\over sin(90^0+\alpha)}\)
From Fig. A, cos α \(={60\over 120}\)
α = 300
\(∴\ P={500\times sin150^0\over sin120^0}\)
A smooth ring A of mass m can slide on a fixed horizontal rod. A string tied to the ring passes over a fixed pulley B and carries a block Cof mass M (2 m) as shown in figure. At an instant, the string between the ring and the pulley makes an angle e with the rod. With what acceleration will the ring start moving, if the system is released from rest with θ = 30°?
- (a)
2.39 rn/s2
- (b)
7.29 rn/s2
- (c)
4.67 rn/s2
- (d)
6.78 rn/s2
Let it starts with acceleration a then
Mg - T = Ma . cos θ ...(i)
T cosθ=ma ...(ii)
By Eqs. (i) and (ii), we get
\(a={Mg\ cos\theta\over m+M\ cos^2\theta}\)
Putting the values
\(a={2mg\ cos30^0\over m+23m\ cos^230^0}=6.78m/s^2\)
A uniform sphere of mass 200 g rolls without slipping on a plane surface so that its centre moves at a speed of 2.00 crn/s. Find its kinetic energy
- (a)
5.6 x 10-5 J
- (b)
3.2 x 10-5 J
- (c)
7.1 X 10-5 J
- (d)
None of these
Kinetic Energy (KE) = \({1\over2}mv^2+{1\over2}l\omega^2\)
\(={1\over2}mv^2+{1\over2}\times{2\over5}mr^2.\omega^2\)
\(={1\over2}mv^2+{1\over5}mv^2={7\over10}mv^2\)
\(={7\over10}\times0.2\times(2\times10^{-2})^2\)
\(={14\over 100}\times4\times10^{-4}\)
= 5.6 X 10-5 J
The 2 kg mass C moving horizontally to the right, with a velocity of 5 mis, strikes the 8 kg mass B at the lower end of the rigid massless rod AB. The rod is suspended from a frictionless hingeatA and is initially at rest. If the coefficient of restitution between mass C and B is one, the angular velocity of rod AB immediately after impact is
- (a)
0.50 rad/s
- (b)
0.75 rad/s
- (c)
0.25 rad/s
- (d)
0.625 rad/s
Initial velocity of mass C = 5 mls = vc
mC = 2 kg
VB = 0 (initially at rest)
mB = 8 kg
Coefficient of restitution = 1
Now,
Relative velocity of approach
= Relative velocity of separation
Approach velocity = vC
Let after impact velocity of C =v'C
and velocity of B = v'B
∴ vC = v'B - v'B
Applying conservation of linear momentum,
mCvC = mBv'B - mCv'C
10 = 4v'B - v'C
or v'B - v'C = 2
or 3v'B = 3
v'B = 1m/s
\(ω={v'\over 1.6}={1\over 1.6}\)
=0.625rad/s
Three particles of masses 0.50 kg, 1.0 kg and 1.5 kg are placed at the three corners of a right-angled triangle of sides 3.0cm, 4.0cm and5.0cm as shown in figure below. Locate the centre of mass of the system.
- (a)
x = 1.3, y = 1.5
- (b)
x=4,y=3
- (c)
x = 0.9, y = 0.7
- (d)
x = 2, y = 1.5
Let us take the 4.0 em line as the X-axis and the 3.0 em line as the Y-axis. The coordinates of the three particles are as follows:
| m | x | y |
|---|---|---|
| 0.50kg | 0 | 0 |
| 1.0kg | 4.0cm | 0 |
| 1.5kg | 0 | 3.0cm |
The x-coordinate of the centre of mass is
\(x={m_1x_2+m_2x_2+m_3x_3\over m_1+m_2+m_3}\)
\(={(0.50)\times0+ (1.0)\times (4.0) + (1.5 )\times 0\over 0.50+ 1.0+ 1.5}\)
\(=
{4\over3}=1.3cm\)
The y-coordinate of the centre of mass is
\(y={m_1y_1+m_2y_2+m_3y_3\over m_1+m_2+m_3}\)
\(={(0.50) \times 0 + (1.0 ) \times 0 + (1.5) (3.0)\over 0.50+1.0+1.5}\)
\(={4.5\over3}=1.5cm\)
- (a)
A shell is fired from a cannon with a speed v at an angle θ with the horizontal direction. At the highest point in its path, it explodes into two pieces of equal mass. One of the pieces retraces its path to the cannon. The speed of other piece immediately after explosion is
- (b)
3v cos θ
- (c)
2v cos θ
- (d)
\({3\over2}v cos θ\)
- (e)
\(\sqrt{3\over2}v cos θ\)
At the highest point only horizontal velocity exists.
Momentum at B before explosion = mv cos θ
∴ Momentum at B after explosion.
\(=\left(m\over 2\right)(-v\ cos\theta)+\left(m\over2\right)\times v'\)
where, v' = velocity of the second piece
Equating momentum before impact and after impact, we get
\(=\left(m\over 2\right)(-v\ cos\theta)+\left(m\over2\right)\times v'\)
v' = 3 v cos θ
A bar AB is hinged to a fixed support at A and the centre of a uniform disc at B. Both hinges are frictionless. At a certain instant, the system is observed to be moving in the horizontal plane, when for the bar has an angular velocity ω while disc has an angular velocity of ω relative to the bar. The bar and the disc are each of mass m. Determine
the angular momentum of the bar about the hinge A
- (a)
\({2\over3}ma^2ω\)
- (b)
\({4\over3}ma^2ω\)
- (c)
\({3\over4}ma^2ω\)
- (d)
zero
Angular momentum of bar about hinge A is
LA = lAB X ω
\(={m_{ab}l^2\over 3}\timesω\)
\(={1\over3}\times m\times (2a)^2\timesω={4\over 3}ma^2ω\)
A spring scale indicates a tension Tin the right hand cable of the pulley system shown in figure. Neglecting the mass of the pulleys and ignoring friction between the cable and pulley. The mass m is
- (a)
\(2T\over g\)
- (b)
\(T(1+e^{4\pi})\over g\)
- (c)
\(4T\over g\)
- (d)
None of these
Applying equilibrium mg = 4T
m=\(4T\over g\)
A mass of 35 kg is suspended from a weightless bar AC, which is supported by a cable CB and a pin atA as shown in the figure. The pin reaction at A on the bar AB are
- (a)
Rx = 343.4 N
Ry= 755.4 N - (b)
Rx = 343.4 N
Ry=0 - (c)
Rx = 755.4 N
Ry= 343.4 N - (d)
Rx =755.3 N
Ry= 0
\(CB=\sqrt{(125)^2+(275)^2}\)
=302
\(cos\theta={125\over 302}\)
θ=65.50
Applying equilibrium,
35 x 9.81 = Tcos 65,50
or T = 830 N
Ry = 0
Resolving the forces horizontally,
T sin 65.5° = Rx
or 830 x 0.91 = Rx
or Rx = 755.3 N
A flat road has a curve segment with a radius of 100 m. While negotiating this curve, a vehicle slipped on its tyres as well as tried to roll over at a particular speed. This speed, assuming a friction coefficient of 0.5, is
- (a)
18 m/s
- (b)
25 m/s
- (c)
32 m/s
- (d)
22 m/s
Let v be the velocity of the vehicle at which both the slip as well as lifting up of a pair of one side wheels takes place.
Centrifugal force \(F_e={mv^2\over R}\)
Vehicle will slip, if this is equal to frictional resistance μmg
\(\mu mg={mv^2\over R}\)
\(v={\sqrt{\mu gR}}=\sqrt{0.5\times9.8\times100}\)
=22m/s
Each of the blocks shown in figure has mass 1kg. The rear block moves with a speed ofl m/s towards the front block kept at rest. The spring attached to the front block is light and has a spring constant 50 N/m. Find the maximum compression of the spring.
- (a)
0.10 m
- (b)
0.20 m
- (c)
0.30 m
- (d)
0.50 m
Momentum equation,
mv = (m+ m) x u
\(⇒u={v\over 2}={2\over 2}=1m/s\)
\({1\over2}mv^2={1\over2}k\times x^2_{max}+{1\over2}mu^2+{1\over 2}mu^2\)
\(⇒{1\over2}\times1\times1(2)^2={1\over2}\times50\times x^2+1\times1^2\)
\(⇒x={1\over5}=0.2m\)
A uniform rod of mass M and length a lies on a smooth horizontal plane. A particle of mass m moving at a speed v perpendicular to the length of the rod, strikes it at a distance a/4 from the centre and stops after the collision. The angular velocity of the rod about its centre just after the collision is
- (a)
\(2mv\over 3Ma\)
- (b)
\(2mv\over Ma\)
- (c)
\(3mv\over Ma\)
- (d)
zero
\(mv=MV\Rightarrow V={m\over M}.v\)
Let A be the centre of the rod when it is at rest. Let AB be the line perpendicular to the plane of the figure. Consider the angular momentum of 'the rod plus the particle' system about AB. Initially, the rod is at rest. The angular momentum of the particle about AB is
L = mv (a/4)
After the collision, the particle comes to rest. The angular momentum of the rod about A is
\(\vec{L}=\vec{L}_{CM}+M\vec r_0\times\vec V\)
As \(\vec r_0||\vec V ,\vec r_0\times \vec V=0\)
Thus, \(\vec L=\vec {L}_{CM}\)
Hence, the angular momentum of the rod about AB is
\(L={I\omega}={Ma^2\over12}\omega\)
Thus, \({mva\over 4}={Ma^2\over 12}\omega\ or\ \omega={3mv\over Ma}\)
The area of moment ofinertia of a square of size 1unit about its diagonal is
- (a)
1/3
- (b)
1/4
- (c)
1/12
- (d)
1/6
The square ABCD has BD as diagonal of length √2 units
The height AE = AD sin 60°
\(=1.{1\over \sqrt2}={1\over\sqrt2}\)unit
As we know, the area of moment of inertia of a triangle about its base is
\(I={bh^3\over 12}\)
∴ Required moment of inertia of square about the diagonal \(=2\times{\sqrt2\over 12}.\left(1\over \sqrt2\right)^3\)
\(={1\over6}.\left(1\over \sqrt2\right)^2\)
\(={1\over 12}unit\)
The ratio of tension on the tight side to that on the slack side in a flat belt drive is
- (a)
proportional to the product of coefficient of friction and lap angle
- (b)
an exponential function of the product of coefficient of friction and lap angle
- (c)
proportional to the lap angle
- (d)
proportional to the coefficient of friction
\({T_1\over T_2}={e^{\mu\theta}}\)
Clearly exponential function.
The figure shows a pin-jointed plane truss loaded at the point M by hanging a mass of 100 kg. The number IN of the truss is subjected to a load of
- (a)
zero
- (b)
490 N in compression
- (c)
981 N in compression
- (d)
981 N in tension
Consider the equilibrium of point L
ΣFx = 0 ⇒ FKL = FML
ΣFy= 0 ⇒FLN=0
As there is no external force acting at point L.
A cylindrical body of cross-sectional area A, height Hand density psis immersed to a depth h in a liquid of density p and tied to the bottom with a string. The tension in the string is
- (a)
ρghA
- (b)
(ρs - ρ) ghA
- (c)
(ρ - ρs) ghA
- (d)
(ρh - ρsH) gA
Let T = tension in the string
⇒ T + mg = Buoyancy force
Mass will be given by
m = ρs H· A (Volume x Density)
⇒ Buoyancy force = T + (ρs H· A) g
This Buoyancy force will be equal to
= ρhAg
So, T + (ρsHA)g = ρhAg
⇒ T = Ag (ρh - ρsH )
Al kg block is resting on a surface with coefficient of friction μ.=0.I. A force of 0.8 N is applied to the block as shown in figure. The friction force is
- (a)
zero
- (b)
0.8 N
- (c)
0.89 N
- (d)
1.2 N
Fmax = μN=0.1 x 1 x 10= IN
But Fexter = 0.8 N < Fmax
∴ Facting = 0.8 N
A shell is fired from a cannon with a velocity v at an angle with the horizontal direction. At the highest point in its path, it explodes into two pieces of equal masses. One of the pieces retraces its path to the cannon. The speed of the other piece immediately after the explosion is
- (a)
\(3v\ cos\theta\)
- (b)
\(2v\ cos\theta\)
- (c)
\({3\over2}v\ cos\theta\)
- (d)
\(v\ cos\theta\)
In the arrangement shown in figure, the ends P and Q of an inextensible string move downwards with uniform speed u. Pulleys A and B are fixed.
Mass M moves upwards with a speed
- (a)
2u cos θ
- (b)
u/cos θ
- (c)
2u / cos θ
- (d)
u cosθ
Let M moves with velocity v.
vcos θ = u
v = u/cos θ
Two identical ladders, each of mass M and length L are resting on the rough horizontal surface as shown in figure. A block of mass m hangs from P. If the system is in equilibrium, then magnitude of frictional force at A is
- (a)
\(f=\left(M+m\over 2\right)g\ sin \theta\)
- (b)
\(f=\left(M+m\over 2\right)g\ cos\theta\)
- (c)
\(f=\left(M+m\over 2\right)g\ tan\theta\)
- (d)
\(f=\left(M+m\over 2\right)g\ cot\theta\)
Free body diagram of the left rod
Free body diagram of right rod
Both the rods are in equilibrium.
i.e., ΣF = 0 and Σ\(\tau\) = 0
For left rod
\(N_2+N=\left(M+{m\over2}\right)g\) ....(i)
N1 = F ...(ii)
For right rod
\(N_2+{mg\over 2}+Mg=N\) ...(iii)
N1 = F ...(iv)
\(N_1+N_2+\left(M+{m\over2}\right)g=\left(M+{m\over2}\right)g\)
\(2N_2=0⇒N_2=0\)
From Eq. (i), \(N=\left(M+{m\over2}\right)g\)
On balancing torque about 0,
\(Mg\left({L\over2}cos\theta\right)+F (Lsin\theta) = NL cos \theta\)
\({Mg\over 2}cos\theta+F\ sin\theta=\left(M+{m\over 2}\right)g\ cos\theta\)
\(F==\left(M+{m\over 2}\right)g\ cos\theta\)
A rod AB of mass M and length L is lying on horizontal frictionless surface. A particle of mass travelling along the surface hits the end A of the rod with a velocity Vo in a direction perpendicular to AB. The collision is completely elastic. After collision the particle comes to rest.
The ratio m/M is
- (a)
1/2
- (b)
1/3
- (c)
2/3
- (d)
1/4
Let v be the velocity of the rod. Then, by conservation of momentum
mvo = Mv ...(i)
By conservation of angular momentum,
\(mv_0{L\over 2}=I\omega\) ...(ii)
Applying Newton's law at the point of collision and in the line of collision
\(1=-{(v_2-v_1)\over (u_2-u_1)}\)
\(-1={-\left(v+{\omega L\over 2}-0\right)\over (0-v_0)}\)
By solving Eqs. (i), (ii) and (iii), we get
\({m\over M}v_0+6{m\over M}.{v_0\over L}.{L\over 2}=v_0\)
\({m\over M}={1\over4}\)
A rectangular rigid fixed block has a long horizontal edge. A solid homogeneous cylinder of radius R is placed horizontally at rest with its length parallel to the edge such that the axis of the cylinder and the edge of the block are in the same vertical plane as shown in given figure.
There is sufficient friction present at the edge so that a very small displacement causes the cylinder to roll-off the edge without slipping. The speed of the centre of mass of the cylinder before leaving contact with the edge is
- (a)
\(\sqrt{3\over 7}gR\)
- (b)
\(\sqrt{4\over 7}gR\)
- (c)
\(\sqrt{5\over 7}gR\)
- (d)
\(\sqrt{2\over 3}gR\)
Applying energy conservation from A to B.
Loss of PE = Gain of KE
\(mgh={1\over2}mv^2+{1\over2}I\omega^2\)
\(mgh={1\over2}mv^2+{1\over2}{mR^2\over2}\times{v^2\over R^2}\)
\(mgR (1 - cos \theta) ={3\over4}mv^2\)
Balancing the force,
\(mg\ cos\theta=N+{mv^2\over R}\) ...(ii)
At the time of leaving contact, N = 0
then, \(mg\ cos\theta={mv^2\over R}\)
or mg R cos θ = mv2
Putting it into Eq. (i),
\(mgR - mv^2 ={3\over 4}mv^2\)
\(v=\sqrt{4\over 7}gR\)
Consider a truss PQR loaded at P with a force F as shown in the figure. The tension in the member QR is
- (a)
0.5F
- (b)
0.63F
- (c)
0.73F
- (d)
0.87F
Let the perpendicular from P meets QR at S, so that
PS = QS = a (let) at ∠45°
Now \(tan 30° ={PS\over SR}={a\over SR}\)
Taking moment about Qs
⇒ Fa - RR X 2.732a = 0
\(⇒ R_R={F\over 2.732}=0.366F\)
∴ RQ = F - 0.366 F = 0.634 F
By method of joints, considering the equilibrium of joint Q, we have
ΣFY = 0
FQF sin 45° = RQ = 0.634 F
⇒ FQF = 0.8966 F
and ΣFx = 0
⇒ FQF cos45° = FQR
\(∴\ F_{QR}=0.8966F\times{1\over \sqrt2}=0.634F\)
If point A is in equilibrium under the action of the applied forces, the values of tension TAB and TAC are respectively
- (a)
520 Nand 300 N
- (b)
300 Nand 520 N
- (c)
450 Nand 150 N
- (d)
150 Nand 450 N
By Larni's theorem, we have
\({T_{AB}\over sin120^0}={T_{AC}\over sin150^0}={600\over sin90^0}\)
\(T_{AB}={600\over 1}\times sin120^0=519.61N\)
=520N
TAC = 600 sin 150° = 300 N
An elevator (lift) consists of the elevator cage and a counter weight of mass m each. The cage and the counter weight are connected by a chain that passes over a pulley. The pulley is coupled to a motor. It is desired that the elevator should have a maximum stopping time of t second from a peak speed v. If the inertia of the pulley and the chain are neglected, the minimum power that the motor must have is
- (a)
\({1\over2} mv^2\)
- (b)
\(mv^2\over 2t\)
- (c)
\(mv^2\over t\)
- (d)
\(2mv^2\over t\)
For both masses,
Initial velocity
v1 = v m/s
Final velocity vt = 0
Initial KEi = \({1\over2}mv^2+{1\over2}mv^2=mv^2\)
Final KEf = 0
Power = Rate of change of KE
\(={dKE\over dt}={KE_i-KE_f\over t}\)
\(={mv^2-0\over t}={mv^2\over t}\)
A 1 kg mass of clay, moving with a velocity of 10 mis, strikes a stationary wheel and sticks to it. The solid wheel has a mass of 20 kg and a radius of 1 m. Assuming that the wheel and the ground are both rigid and that the wheel is set into pure rolling motion, the angular velocity of the wheel immediately after the impact is approximately
- (a)
zero
- (b)
\({1\over3}rad/s\)
- (c)
\(\sqrt{10\over3}rad/s\)
- (d)
\({10\over3}rad/s\)
According to the law of conservation of linear momentum.
m1u1 + m2u2 = m1v1 + m2v2
1 x 10 + 20 x 0 =(m1 + m2) v
(u2 = 0 as wheel is at rest and v1 = v2 = v)
v (21) = 10
\(v={10\over21}m/s\)
But v=r.ω
∴ Angular velocity \(ω={v\over r}\)
\(={10\over 21\times41 }={10\over 21}rad/s\)
= 0.476 rad/s
which is approximately near to 1/3 rad/s
An ejector mechanism consists of a helical compression spring having a spring constant ofk= 981 x 103 N/m.lt is pre-compressed by 100 mm from its free state. If it is used to eject a mass of 100 kg held on it, the mass will move up through a distance of
- (a)
100 mm
- (b)
500 mm
- (c)
981 mm
- (d)
1000 mm
Spring rate k = 981 X 103 N/m
= 981 N/mm
Mass kept on it, m= 100 kg
∴ Force exerted on spring F = mg
= 100 x 9.81 = 981 N
So, deflection in spring
\(\delta={F\over k}={981\over 981}=1mm\)
∴ The mass will be lifted upto the extent of pre-compression i.e, 100 mm.
A reel of mass m and radius of gyration k is rolling down smoothly from rest with one end of the thread wound on it held in the ceiling as depicted in the figure. Consider the thickness of the thread and its mass negligible in comparison with radius r of the hub and the reel mass m. Symbol g represents the acceleration due to gravity.
The tension in the thread is
- (a)
\(mgr^2\over (r^2+k^2)\)
- (b)
\(mgrk^2\over (r^2+k^2)\)
- (c)
\(mgk^2\over (r^2+k^2)\)
- (d)
\(mg\over (r^2+k^2)\)
Continuing from the previous solution
\(T={mk^2\over r^2}a\)
\(T={mk^2\over r^2}\times {gr^2\over (k^2+r^2)}\)
\(T={mgk^2\over k^2+r^2}\)
Bodies 1 and 2 shown in the figure have equal mass m. All surfaces are smooth. The value of force P required to prevent sliding of body 2 on body 1 is
- (a)
P= 2 mg
- (b)
P= √2mg
- (c)
P= 2√2mg
- (d)
P= mg
By considering the FBD of body 2,
N = mg cos 45°
There is no frictional force as the surfaces are smooth.
The downward force along the slide mg sin 45° is to be balanced by P cos 45°.
\(P\times {1\over \sqrt2}=mg={1\over \sqrt2}\)
⇒ P=mg
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