Q#68
A meter stick weighing 240 g is pivoted at its upper end in such a way that it can freely rotate in a vertical plane through this end (figure 10E-12). A particle of mass 100 g is attached to the upper end of the stick through a light string of length 1 m. Initially, the rod is kept vertical and the string horizontal when the system is released from the rest. The particle collides with the lower end of the stick and sticks there. Find the maximum angle through which the stick will rise.
Answer:
Mass of the stick M = 240 g = 0.24 kg
Mass of the particle m = 100 g = 0.10 kg
Let the angular speed of the particle with string = ⍵
M.I. = I = ml² = 0.10(1)² =0.10 kg.m²
The rotational energy of the string particle = P.E.
½I⍵² = mgl
0.5(0.10)⍵² = 0.10g(1)
⍵² = g/0.5 = 2g
⍵ = √(2g)
The angular momentum of the system before the strike
= Angular momentum of the string-particle+Angular momentum of the rod
= I⍵ + 0 = I⍵
= 0.10⍵ = 0.10√2g
Let the angular speed of the rod after the strike be ⍵'
Now the angular momentum = I'⍵'
= (Ml²/3 + ml²)⍵'
= (0.24/3 + 0.10)⍵'
= 0.18⍵'
The angular momentum will be conserved.
0.18⍵' = 0.10√(2g)
⍵' = 10√2√g/18 = 0.786√g
KE = ½I'⍵'² = 0.5(0.18)⍵'² = 0.09(0.786)0.786g = 0.055 g (i)
the change in PE of the system at angle θ after the strike
= mg(1 – cosθ) + Mg(½ – ½cosθ)
= 0.10g(1 – cosθ) + 0.12g(1 – cosθ)
= (1 – cosθ)(0.22g)
Equating (i) and (ii)
(1 – cosθ)(0.22g) = 0.055g
1 – cosθ = 0.055/0.22 = 0.252
cosθ = 1 – 0.252 = 0.747
θ ≈ 41°
Q#69
A uniform rod pivoted at its upper end hangs vertically. It is displaced through an angle of 60° and then released. Find the magnitude of the force acting on a particle of mass dm at the tip of the rod when the rod makes an angle of 37° with the vertical.
Answer:
Let the mass of the rod be m and length l. If the angular speed of the rod at 37° be ⍵, then change in K.E. of the rod should be equal to change in P.E. of the rod.
Change in K.E. = ½I⍵² – 0 = ½(ml²/3)⍵² = ml²⍵²/6
Change in P.E. = mg(vertical distance of the center of the rod at 37°-vertical distance of the center of the rod at 60°)
= ½mgl(cos37°– cos60°)
= ½mgl(0.80 – 0.5)
= 0.15mgl
Equating we get,
ml²⍵²/6 = 0.15mgl
⍵² = 0.9g/l
Let the angular acceleration at 37° be α. Equating the torque at this point,
Iα = mgsin37°(½l)
{mgsin37° is the component of the weight of the rod perpendicular to it and acting at the center of the rod. Multiplying it by l/2 we get the torque about the pivot}
(ml²/3) α = ½(0.60)mgl = 0.30mgl
α = (3/l)(0.30g)
α = 0.90g/l
There will be tangential and radial forces on the particle dm at the tip of the rod.
Radial force Fₓ = dm(⍵²l) = dm(0.9g/l)(l) = 0.9gdm
Tangential force Fᵧ = mass x acceleration = dm (αl) = dm(l)(0.90g/l) = 0.9dm g = Fₓ
These two forces are perpendicular to each other.
Hence the resultant force on the particle dm,
F = √(Fₓ² +Fᵧ²) = √2Fₓ
= (√2)0.90dm(g)
= 0.9g√2dm
Q#70
A cylinder rolls on a horizontal plane surface. If the speed of the center is 25 m/s, what is the speed of the highest point?
Answer:
When a cylinder or wheel rolls on a surface the speed of the point in contact is zero and its vertical diameter can be considered as rotating about the point of contact. Let the angular speed of the vertical diameter about the point of contact be ⍵.
⍵ = v/r = 25/r rad/s. If the speed of the highest point be V, then
V = ⍵(2r) = (25/r)(2r) =50 m/s
Q#71
A sphere of mass m rolls on a plane surface. Find its kinetic energy at an instant when its center moves with speed v.
Answer:
At this stage, the sphere will have linear as well as rotational kinetic energy.
K.E. = ½mv² + ½I⍵²
I = 2mr²/5
⍵ = v/r
Hence total K.E. = ½mv² + ½(2mr²/5)(v²/r²)
= ½mv² + mv²/5
= 7mv²/10
Q#72
A string is wrapped over the edge of a uniform disc and the free end is fixed with the ceiling. The disc moves down, unwinding the string. Find the downward acceleration of the disc.
Answer:
Let the acceleration of the disc be a and the tension in the string T. If the radius of the disc is r, then torque on the disc = Tr.
Now, mg – T = ma
T = mg – ma
The angular acceleration of the disc = α = a/r
Torque = Iα
Tr = (½mr²)(a/r)
2T = ma
2mg – 2ma = ma
2g = 3a
a = 2g/3
Q#73
A small spherical ball is released from a point at a height h on a rough track shown in figure (10-E13). Assuming that it does not slip anywhere, find its linear speed when it rolls on the horizontal part of the track.
Answer:
The P.E. of the ball = mgh will be converted to K.E. of the ball when it rolls on the horizontal part of the track.
Let the linear speed of the ball be v. Translational K.E. = ½mv²
The angular speed of the ball ⍵ = v/r
and the moment of inertia of the ball, I = 2mr²/5
The angular K.E.
= ½I⍵² = ½(2mr²/5)⍵² = mr²⍵²/5
Equating the total K.E. and the P.E.
½mv² + mr²⍵²/5 = mgh
5v²+2r²(v/r)² = 10gh
7v² = 10gh
v² = 10gh/7
v = √(10gh/7)
T = mg – ma
The angular acceleration of the disc = α = a/r
Torque = Iα
Tr = (½mr²)(a/r)
2T = ma
2mg – 2ma = ma
2g = 3a
a = 2g/3
Q#73
A small spherical ball is released from a point at a height h on a rough track shown in figure (10-E13). Assuming that it does not slip anywhere, find its linear speed when it rolls on the horizontal part of the track.
Answer:
The P.E. of the ball = mgh will be converted to K.E. of the ball when it rolls on the horizontal part of the track.
Let the linear speed of the ball be v. Translational K.E. = ½mv²
The angular speed of the ball ⍵ = v/r
and the moment of inertia of the ball, I = 2mr²/5
The angular K.E.
= ½I⍵² = ½(2mr²/5)⍵² = mr²⍵²/5
Equating the total K.E. and the P.E.
½mv² + mr²⍵²/5 = mgh
5v²+2r²(v/r)² = 10gh
7v² = 10gh
v² = 10gh/7
v = √(10gh/7)
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