a thin and uniform rod of mass M and length L is held vertical on a floor with large friction . the rod is released from rest so that it falls by rotating about its contact point with the floor without slipping. which of the following statement is correct when the rod makes an angle 60 with vertical
Answers
Given: uniform rod of mass M and length L, the rod makes an angle 60 with vertical.
To find: normal, radial acceleration and angular speed.
Solution:
- Very first, lets make the sum of kinetic energy and potential energy to be zero, we get:
ΔK + ΔU = 0
where ΔK is kinetic energy and ΔU be the potential energy.
ΔK = -ΔU
1/2 x I x ω² = -ΔU
I is moment of inertia of the rod = mL²/3
1/2 x mL²/3 x ω² = - (-mgL/4)
ω = √3g/2L angular speed
- Now for radial acceleration:
a(radial) = ω²L/2
a(radial) = 3gL/2L x L/2
a(radial) = 3g/4
- Now,
α= (mgL/2 x sin 60°)/ (mL²/3)
α = 3√3g/4L
- Now, vertical acceleration can be calculated as:
a(vertical) = (αL/2) x sin 60° + ω²L/2 x cos 60°
- So, after calculating, vertical acceleration comes out to be :
a(vertical) = 9g/16 + 6g/16
a(vertical) = 15g/16
- Now, all the vertical forces is equal to ma(vertical), so:
mg - N = ma(vertical)
N = mg/16
Answer:
So the normal reaction is: mg/16 , radial acceleration is 3g/4 and angular speed is: √3g/2L
Complete question:
A thin and uniform rod of mass M and length L is held vertical on a floor with large friction. The rod is released from rest so that it falls by rotating about its contact-point with the floor without slipping. Which of the following statement(s) is/are correct, when the rod makes an angle 60° with vertical? [g is the acceleration due to gravity]
A. The angular acceleration of the rod will be 2g/L
B. The normal reaction force from the floor on the rod will be Mg/16
C. The radial acceleration of the rod's center of mass will be 3g/4
D. The angular speed of the rod will be √(3g/2L)
Answer:
The correct statement is B. The normal reaction force from the floor on the rod will be Mg/16
Explanation:
On applying conservation of energy, we get,
ΔK + ΔU = 0 → (equation 1)
The change in kinetic energy is given as:
ΔK = 1/2 I₀ ω²
⇒ I₀ = mL²/3
The change in potential energy is given as:
ΔU = -mg × L/4
On substituting ΔK and ΔU on equation 1, we get,
(1/2 I₀ ω²) + (-mg × L/4) = 0
1/2 × mL²/3 × ω² = mg × L/4
1/2 × L/3 × ω² = g × 1/4
ω² = 3/2 × g/L
∴ ω = √(3g/2L)
The radial acceleration is given as:
Ar = ω² × L/2
On substituting the value of 'ω' in above equation, we get,
∴ Ar = 3g/2L × L/2 = 3g/4
Now,
τ = αI₀
α = (mgL/2 x sin 60°)/(mL²/3)
α = 3√3g/4L
The vertical acceleration is given as:
Av = ((αL)/2 × sin 60°) + ((ω²L)/2 × cos 60°)
Av = 9g/16 + 6g/16
∴ Av = 15g/16
Net force = Vertical acceleration
mg - N = m × Av
N = mg - (m × 15g/16)
N = (16mg - 15mg)/16
∴ N = mg/16