Q#63
Two small balls A and B, each of mass m, are joined rigidly to the ends of a light rod of length L (figure 10E-10). The system translates on a frictionless horizontal surface with a velocity v₀ in a direction perpendicular to the rod. A particle P of mass m kept at rest on the surface sticks to the ball A as the ball collides with it. Find (a) the linear speeds of the balls A and B after the collision, (b) the velocity of the center of mass C of the system A + B + P and (c) the angular speed of the system about C after the collision.
Answer:
(a) Let the velocity of A after the strike = v'
Since the light rod will exert a force only along its length, the strike with the particle A will not affect the velocity of B. So, equating the momentum of A before and after the collision.
mv₀ = 2mv'
v' = v₀/2 (Velocity of A after the collision)
The velocity of B = v₀
(b) The velocity of the center of mass A + B + P = V
3mV = [2m(v₀/2) + m(v₀)]
3mV = 2mv₀
V = 2v₀/3
{Along the initial direction of the velocities.}
(c)Let the angular speed of the system about C be ⍵.
The distance of CoM from A = x = mL/3m = L/3
and from B = 2L/3.
Let us consider the velocities of A and B with respect to CoM.
The velocity of A w,r,t CoM = v₀/2 – 2v₀/3
= (3v₀ – 4v₀)/6
= –v₀/6 {negative sign says the velocity of A is towards left with respect to CoM}
Hence, ⍵ = Velocity/radius
= (v₀/6)/(L/3) = (v₀/6)(3/L)
= v₀/2L (Clockwise).
It can be confirmed with respect to particle B also. Velocity of B w.r.t. CoM
= v₀ – 2v₀/3 =v₀/3 {Towards right}
Hence ⍵ = (v₀/3)/(2L/3) =(v₀/3)(3/2L) = v₀/2L (Clockwise)
Q#64
The rod with the balls A and B of the previous problem are clamped at the center in such a way that it rotates freely about a horizontal axis through the clamp. The system is kept at rest in the horizontal position. A particle P of the same mass m is dropped from a height h on the ball B. The particle collides with B and sticks to it. (a) Find the angular momentum and the angular speed of the system just after the collision. (b) what should be the minimum value of h so that the system makes a full rotation after the collision.
Answer:
(a) The speed of the particle P when it reaches ball B = v = √(2gh)
Hence the angular momentum of the particle P about the horizontal axis (center of the rod) = mass x velocity x radius
L = m(√(2gh))(L/2)
L = mL√(gh)/√2
Since there is no external force on the system, this angular momentum will be conserved even after the strike.
Let the angular speed of the rod after the strike be ⍵. Now the moment of inertia I about the center
I = 2m(L/2)² + m(L/2)²
I = 3mL²/4
The angular momentum
L = I⍵ = 3mL²⍵/4
Equating this with the angular momentum calculated above, we get
3mL²⍵/4 = mL√(gh)/√2
⍵ = (4/3L)(√(gh)/√2)
= 4√2√(gh)/6L {Multiplying num. and den. by √2}
= √(8gh)/3L
(b) The kinetic energy after the strike
K.E = ½I⍵²
K.E = ½(3mL²/4){√8gh)/3L}²
K.E = (3mL²/8)(8gh/9L²)
K.E = mgh/3
Let us take the horizontal axis as a reference for P.E. The system will just complete a full rotation if the ball B with particle P just reaches the topmost position. Here the K.E. will be zero and fully converted to P.E. At this position the P.E.
= 2mg(L/2) – mg(L/2)
= mgL/2
Equating K.E. and P.E.
mgh/3 = mgL/2
h/3 = L/2
h = 3L/2
Q#65
Two blocks of masses 400g and 200 g are connected through a light string going over a pulley which is free to rotate about its axis. The pulley has a moment of inertia 1.6 x 10⁻⁴ kg.m² and a radius 2.0 cm. Find (a) the kinetic energy of the system as the 400 g block falls through 50 cm, (b) the speed of the blocks at this instant.
Answer:
(a) M = 400 g = 0.40 kg
m = 200 g = 0.20 kg
the fall of mass 400 g = h = 50 cm =0.50 m
Drop in P.E. of the 400 g = mgh = 0.40(9.8)0.50 J
= 1.96 J
Simultaneously the 200 g mass will rise through 0.50 m, hence rise in P.E. of 200 g
=0.20(9.8)0.50 J = 0.98 J
Net loss of P.E. = 1.96 – 0.98 = 0.98 J
Hence eaqual amout will convert to K.E. = 0.98 J
(b) I = 1.6 x 10⁻⁴ kg.m², r = 2.0 cm = 0.02 m
Let the speed of the blocks at this instant be = v
The angular speed of the pulley = ⍵ = v/r
Total K.E. of the system = K.E. of the 400 g mass + K.E. of the 200 g mass + K.E. of the pulley
= ½(0.40)v² + ½(0.20)v² + ½I⍵²
= 0.20v² + 0.10v² + ½(1.6 x 10⁻⁴)(v/0.02)²
= 0.30v²+ (1.6 x 10⁻⁴ v²/0.0008) J
= 0.50v² J
But it has been calculated to be 0.98 J in part (a) of the problem. Equating,
0.50v² = 0.98
v² = 0.98/0.50
v = 1.4 m/s
Q#66
The pulley shown in the figure (10E-11) has a radius of 20 cm and a moment of inertia 0.20 kg-m². The string going over it is attached at one end to a vertical spring of spring constant 50 N/m fixed from below and supports a 1 kg mass at the other end. The system is released from rest with the spring at its natural length. Find the speed of the block when it has descended through 10 cm. Take g = 10 m/s².
Answer:
Let the speed of the block be v when it descends through 10 cm. At this time the spring is also stretched through 10 cm. So the force on the spring = kx = 50(10/100) = 5 N
So the P.E. stored in the spring = ½kx² = 0.5(50)(0.10)0.10 = 0.25 J
Radius of the pulley, r = 20 cm = 0.20 m.
Moment of inertia of the pulley = I = 0.20 kg-m².
The angular speed of the pulley, ⍵ =v/r = v/0.20
K.E. of the pulley
= ½I⍵² = ½(0.20)v²/(0.20)² = v²/0.4
K.E. of the block = ½mv² = 0.50(1)v² = 0.50 v² J
Loss of P.E. of the block = mgh = 1(10)0.10 = 1 J
Loss of P.E. of the block = Energy stored in the spring + K.E. of the pulley + K.E. of the block
1 = 0.25 + v²/0.4 + 0.50v²
2.5 v² + 0.50 v² = 0.75
3v² = 0.75
v² = 0.25
v = 0.50 m/s
Q#67
A meter stick is held vertically with one end on a rough horizontal floor. It is gently allowed to fall on the floor. Assuming that the end at the floor does not slip, find the angular speed of the rod when it hits the floor.
Answer:
The height of the center of mass of the stick from the floor = 0.50 m
P.E. of the stick in the vertical position = mgh = 0.50mg J
When it hits the floor its P.E. becomes zero and it is converted to K.E. Let the angular speed of the stick when it hits the ground = ⍵
K.E. of the stick = ½I⍵²
Since the stick rotates about the end, the M.I. about the end
= I₀ + mr²
= ml²/12 + m(l/2)²
= ml²/12 + ml²/4
= 4ml²/12
= ml²/3
Now K.E. of the stick = ½(ml²⍵²/3) J
= ml²⍵²/6 J
K.E. = P.E.
ml²⍵²/6 = 0.50mg
⍵² = 3g/l²
= 3(9.8)/12
⍵ = 5.4 rad/s
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