Q#13
A simple pendulum is suspended from the ceiling of a car taking a turn of radius 10 m at a speed of 36 km/h. Find the angle made by the string of the pendulum with the vertical if this angle does not change during the turn. Take g = 10 m/s².
Answer:
Let the mass of the bob be = m
Speed of the car = 36km/h = 36000/3600 m/s
= 10 m/s
The radius of the turn, r = 10 m
The horizontal force (Outward)on the bob at the turn = mv²/r
= m100/10 = 10m
Vertical force (Downward) on the bob
= Weight of bob, mg = 10m
Resultant of these two forces is balanced by the tension in the string. So, the resultant and the string is in the same line. See the diagram below,
Let the string make an angle θ with the vertical. So, the resultant of these two forces also make an angle θ with the downward force.
tanθ = 10m/10m = 1
Ө = 45°
Q#14
The bob of a simple pendulum of length 1 m has mass 100 g and a speed of 1.4 m/s at the lowest point in its path. Find the tension in the string at the instant.
Answer:
Tension in the string at the lowest point will be due to the weight of the bob and circular motion of the bob.
i.e. T = mg + mv²/r
Here m = 100 g = 0.10 kg, v =1.4 m/s, r = 1 m
So,
T = 0.10 x 10 + 0.10 x 1.4²/1
= 1 + 0.196 N ≈ 1.2 N
Q#15
Suppose the bob of the previous problem has a speed of 1.4 m/s when the string makes an angle of 0.20 radian with the vertical. Find the tension at the instant. You can use cosθ ≈ 1 – θ²/2 and sinθ ≈ θ for small θ.
Answer:
Here θ = 0.20 radian. Radial outward force on the bob due to circular motion
= mv²/r = 0.10(1.4)²/1 = 0.196 N.
Component of weight along the string line = mg.cosθ
= 0.10 x 9.8 (1 – θ²/2) = 0.98(1 – θ²/2)
= 0.98(1 – 0.02) = 0.96 N
Total outward force on the bob along the string line
= 0.196 + 0.96 = 1.156 N ≈ 1.16 N
This outward force on the string is balanced by the tension in the string which is also equal to 1.16 N.
Q#16
Suppose the amplitude of a simple pendulum having a bob of mass m is θ. Find the tension in the string when the bob is at its extreme position.
Answer:
Tension T in the string is equal to the total radial force on the bob at the instance. So,
T = mv²/r + mgcosθ, (See figure below),
where v is the speed of the bob. The first term is the force due to circular motion while the second term is the component of the weight along the radial line. In the given problem the bob is at its extreme position so v = 0.
T = mg.cosθ
Q#17
A person stands on a spring balance at the equator. (a) By what fraction is the balance reading less than his true weight? (b) If the speed of the earth rotation is increased by such an amount that the balance reading is half the true weight, what will be the length of the day in this case?
Answer:
(a) Let time period of the earth for one rotation around its own axis, T = 24 hrs.
The angular speed of the earth ω = 2π/T = 2π/(24 x 3600) radians/s
The outward force due to it = mω²R (R – radius of the earth, taking it = 6400 km)
It is this outward force by which the true weight of the person reduces.
The fraction by which the weight reduces = mω²R/mg
= ω²R/g = 4π² x 6400 x 1000/(24 x 24 x 3600 x 3600 x 9.8)
= 80π²/(6 x 3 x 36 x 36 x 9.8) = 3.45 x 10-3
(b)In this case the outward force due to the rotation of earth = Half the true weight
mω²R = ½mg
2(2π/T)²R = g
8π²R = gT²
T² = 8π²R/g = 8 x 9.86 x 6400 x 1000/9.8 = 51513469.4
T = 7177 s =7177/3600 hr = 1.994 hr ≈ 2 hrs
Q#18
A turn of radius 20 m is banked for the vehicles going at a speed of 36 km/h. If the coefficient of static friction between the road and the tyre is 0.4, what are the possible speeds of a vehicle so that it neither slips down nor skids up?
Answer:
If the vehicle is going at a speed of 36 km/h the equilibrium of forces is as in figure below,
So, N.cosθ = mg and N.sinθ = mv²/r
Dividing later by former, we get
tanθ = v²/gr
Here r = 20 m, v = 36000/3600 m/s = 10 m/s
So, tanθ = 100/10 x 20 = 1/2
(Taking g = 10 m/s²)
But it is the case where friction does not come into the picture.
Let the maximum velocity without skidding up be V. Now a frictional force, F will act on the vehicle along the plane as shown in the picture below.
F = µN
Now, Nsinθ + Fcosθ = mV²/r (Horizontal equilibrium of forces)
Nsinθ = mV²/r – Fcosθ (i)
Ncosθ – Fsinθ = mg (Vertical equilibrium of forces)
Ncosθ = mg + Fsinθ (ii)
Dividing (i) by (ii) we get,
tanθ = (mV²/r – Fcosθ)/(mg + Fsinθ) = 1/2
(we got tanθ = 1/2, so sinθ = tanθ/√(1 + tan²θ) = 1/√5
and cosθ = 1/√(1 + tan²θ) = 2/√5)
mg + Fsinθ = 2mV²/r – 2Fcosθ
F(sinθ + 2cosθ) = m(2V²/r – g)
µN(1/√5 + 2 x 2/√5) = m(2V²/20 – 10)
0.4 N(5/√5) = m(V² – 100)/10
4√5 N = m(V² – 100) (iii)
Resolving forces perpendicular to surface, we get
mgcosθ + (mV²/r)sinθ = N
N = 10m(2/√5) + mV²/20√5 = m(20/√5 + V²/20√5)
N = m(400 + V²)/20√5
20√5N = m(400 + V²) (iv)
Dividing (iii) b(iv) we get,
1/5 = (V² - 100)/(V² + 400)
V² + 400 = 5V² - 500
4V² = 900
V² = 900/4
V = 30/2 = 15 m/s = 15 x 3600/1000 = 54 km/hr
Now let us consider the case for minimum velocity 'u' on the turn so that vehicle does not slip down.
In this case, the force of friction will act upward along the plane. See the picture below:
N sinθ = F cosθ + mu²/r (v)
(Horizontal equilibrium of forces)
Ncosθ + Fsinθ = mg
(Vertical equilibrium of forces)
N.cosθ = mg – Fsinθ (vi)
Dividing (v) by (vi) we get,
tanθ = (Fcosθ + mu²/r)/(mg – F sinθ) = 1/2 (Since tan θ = 1/2)
(mg – Fsinθ) = 2Fcosθ + 2mu²/r
F(sinθ + 2cosθ) = m(g – 2u²/r)
µN(1/√5 + 2 x 2/√5) = m(10 – 2u²/20)
0.4N√5 = m(100 – u²)/10
4√5N = m(100 – u²) (vii)
Now resolving the forces along N,
N = mg cosθ + (mu²/r)sinθ = m [gcosθ + (u²/r) sinθ]
N = m(10 x 2/√5 + u²/20√5)
N = m(u² + 400)/20√5
20√5N = m(u² + 400) (viii)
Dividing (vii) by (viii), we get
1/5 = (100 – u²)/(u² + 400)
u² + 400 = 500 – 5u²
6u² = 100
u² = 100/6
u = 10/√6 m/s = 10 x 3600/√6 x 1000 km/hr
= 36/√6 km/hr = 14.7 km/hr
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