Q#19
A motorcycle has to move with a constant speed on an over-bridge which is in the form of the circular arc of radius R and has a total length L. (a) Suppose the motorcycle starts from the highest point. What can its maximum velocity be for which the contact with the road is not broken at the highest point? (b) If the motorcycle goes at speed 1/√2 times the maximum found in part (a), where will it lose the contact with the road? (c) what maximum uniform speed can it maintain on the bridge if it does not lose contact anywhere on the bridge?
Answer:
Due to circular motion, the motorcycle will experience a radial force = mv²/R, where m is mass of motorcycle and v is the velocity.
(a) At the highest point, this force is upwards while the weight of the motorcycle is downward. So not to break contact with the road the maximum velocity should be such that
mv²/R = mg
v² = Rg
v = √(Rg)
(b) Now v = 1/√2 x √(Rg) = √(Rg/2)
Let at an angular distance θ from the vertical it is about to lose the contact. It is clear that the contact-making force i.e. weight 'mg' is now not in line with the contact breaking force mv²/R. Only a component of mg will counter it. Equilibrium of the forces in the radial direction is,
mv²/R = mgcosθ
v²= Rgcosθ
(√(Rg/2)² = Rgcosθ
Rg/2 = Rg.cosθ
cosθ = 1/2
θ = 60° = π/3 Radian
This point will be at a distance l=Rθ from the highest point along the bridge.
i.e. l = Rπ/3
(c) As we have seen above in (b) that v² = Rgcosθ
In this problem, Rg is constant, so as θ increases cosθ decreases and so is v. It means at the start or end of the overbridge where θ is maximum, the motorcycle will first lose contact.
Distance along the bridge from the top at this point = L/2
so θ = (L/2)/R = L/2R
Putting this value in v² = Rgcosθ
v² = Rg cos(L/2R)
v = √{Rgcos(L/2R)}
Q#20
A car goes on a horizontal circular road of radius R, the speed increasing at a constant rate dv/dt = a. The friction coefficient between the road and the tyre is µ. Find the speed at which the car will skid.
Answer:
Since the circular motion is non-uniform, the car will have two-component of acceleration, Radial and tangential.
Magnitude of Radial acceleration = v²/R
Magnitude of Tangential acceleration = dv/dt = a
Magnitude of total acceleration = √{(v²/R)² + a²}
Horizontal Force on the car, P = m√{(v²/R)² + a²}
Normal Force on the car, N = mg
Limiting Frictional force on the car, F = µN = µmg
Just when the car is about to skid, P = F
m√{(v²/R)² + a²} = µmg
√{(v²/R)² + a²} = µg
(v²/R)² + a² = µ²g²
(v²/R)² = µ²g²-a²
v⁴/R² = µ²g²-a²
v⁴ = (µ²g² - a²)R²
v = {(µ²g² - a²)R²}¹/⁴
Q#21
A block of mass m is kept on a horizontal ruler. The friction coefficient between the ruler and the block is µ. The ruler is fixed at one end and the block is at a distance L from the fixed end. The ruler is rotated about the fixed end in the horizontal plane through the fixed end. (a) What can the maximum angular speed be for which the block does not slip? (b) If the angular speed of the ruler is uniformly increased from zero at an angular acceleration α, at what angular speed will the block slip ?
Answer:
(a) Let Max Angular Speed be ω. At this angular speed the block will experience an outward force F= mω²L. This will be just equal to the frictional force = µmg. Equating the two we get,
mω²L = µmg
ω² = µg/L
ω = √(µg/L)
(b) When the speed is increased with an angular acceleration, the circular motion becomes non-uniform. So the block will have both radial and tangential accelerations.
The radial acceleration will be same as above = ω²L.
The tangential acceleration = dv/dt = d(ωL)/dt = Ldω/dt = Lα
So resultant acceleration will be
= √{(ω²L)² + (Lα)²}
= L√(ω4 + α²)
So the force on it will be, mL√(ω4 + α²) = µmg
ω4 + α² = µ²g²/L² (Squaring both sides)
ω = (µ²g²/L² - α²)1/4
Q#22
A track consists of two circular parts ABC and CDE of equal radius 100 m and joined smoothly as shown in figure (7-E1). Each part subtends a right angle at its centre. A cycle weighing 100 kg together with the rider travels at a constant speed of 18 km/h on the track. (a) Find the normal contact force by the road on the cycle when it is at B and at D. (b) Find the force of friction exerted by the track on the tyres when the cycle is at B, C and D. (c) Find the normal force between the road and the cycle just before and just after the cycle crosses C. (d) What should be the minimum friction coefficient between the road and the tyre, which will ensure that the cyclist can move with constant speed. ? Take g = 10 m/s².
Answer:
(a) B and D are crest and trough of the curvilinear road where the normals to the surface are vertical. So normal contact forces at these points by the road on the cycle will be vertically upwards balancing the effective weight of the cycle along with the rider.
Since the cycle has a uniform circular motion in the vertical plane, it will experience a radially outward force = mv²/r. At B its direction will be vertically upward opposite to weight. Thus effective weight of the cycle along with the rider at B
= mg – mv²/r
(m = 100 kg, g = 10m/s², r = 100 m & v = 18 km/h = 18000/3600 m/s = 5m/s)
= 100 x 10 – 100 x 5²/100 N
= 1000 – 25 N = 975 N = Normal Contact force by the road on the cycle at B.
Effective weight of the cycle and rider at D
= mg + mv²/r
= 100 x 10 + 100 x 5²/100 N
= 1000 + 25 N
= 1025 N = Normal contact force by the road on the cycle at D.
(b) At B and D the weight is balanced by the Normal Contact force and no force is on the cycle along the road. So there is no force of friction to oppose it at B and D. i.e. F = 0 N
At C the weight is acting at an angle of 45° to the surface. Resolving it along the surface and perpendicular to it we get a force equal to mg cos45° acting on the cycle to accelerate it along the track. But the cycle does not accelerate, only goes along the track with a constant speed. So the force of friction is also equal to mg cos45° (opposite to the direction of motion)
= 100 x 10 x 1/√2 = 500√2 N = 707 N
(C) Normal force near the point C will be equal to the component of weight perpendicular to the surface = mg sin45°
= 100 x 10 x 1/√2 = 500√2 N =707 N
Magnitude of Radial force just before and after the point C
= mv²/r = 100 x 5²/100 = 25 N
But before C due to convex nature of the surface, its direction will be opposite to weight component and reducing it to 707 – 25 N
= 682 N = Normal force.
And after C due to the concave nature of the surface its direction will be along the weight component, thus total push on the road will be equal to 707 + 25 N = 732 N. Therefore Normal force here will also be = 732 N
(d) Let the minimum coefficient of friction be µ.
At C, Magnitudes of Frictional force is equal to = 707 N,
Normal forces just before and after C are 682 N and 732 N respectively. So coefficient of friction µ at these points are 707/682 = 1.037 N and 707/732 = 0.96
Though the lower value of µ is 0.96 but if we take this value just before C this will make available frictional force less than 707 N while the tangential component of weight remains 707 N. Thus it will skid and its speed won't be constant. So minimum coefficient of friction should be µ = 1.037
Q#23
In a children's park, a heavy rod is pivoted at the centre and is made to rotate about the pivot so that the rod always remains horizontal. Two kids hold the rod near the ends and thus rotate with the rod (Figure7-E2). Let the mass of each kid be 15 kg the distance between the points of the rod where the two kids hold it be 3.0 m and suppose that the rod rotates at the rate of 20 revolutions per minute. Find the force of friction exerted by the rod on one of the kids.
Answer:
Each kid needs and inward force to remain in uniform circular motion with the horizontal rod and this inward force is provided by the force of friction on the kid by the rod.
Value of this frictional force 'F' is, mω²r.
Here m = 15 kg, r = 3/2 m = 1.5 m,
Ω = 20 x 2π/60 rads/s = 2π/3 rads/s
Now F = 15 x (4π²/9) x 1.5 = 10π² N.
Q#24
A hemispherical bowl of radius R is rotated about its axis of symmetry which is kept vertical. A small block is kept in the bowl at a position where the radius makes an angle θ with the vertical. The block rotates with the bowl without any slipping. The friction coefficient between the block and the bowl surface is µ. Find the range of the angular speed for which the block will not slip.
Answer:
Let the Angular Speed be ω, and the mass of the block be 'm'. See picture below,
For the circular motion of block, radius will be = Rsinθ
Outward horizontal force on the block F = mω²R sinθ
Its tangential component = F cosθ = mω²R sinθ cosθ
Normal component = F sinθ = mω²R sin²θ
Weight of the block = mg
Tangential component of weight = mgsinθ
Normal component of weight = mgcosθ
CASE-I
For minimum angular speed,
Net tangential force = mg sinθ – mω²R sinθ cosθ
Net Normal force = mω²Rsin²θ + mgcosθ
For the condition that block should not slip down,
mg sinθ – mω²R sinθ cosθ = µ(mω²Rsin²θ + mgcosθ)
mg sinθ - µmgcosθ = mω²Rsinθ(cosθ + µsinθ)
g.sinθ-µg.cosθ = ω²R.sinθ.(cosθ+µ.sinθ)
ω² = g(sinθ - µ cosθ)/Rsinθ(cosθ + µsinθ)
ω = √[g(sinθ - µcosθ)/Rsinθ(cosθ + µsinθ)]
CASE-II
For maximum angular speed
Net tangential force = mω²Rsinθcosθ - mgsinθ
Net Normal force = mω²Rsin²θ + mgcosθ
For the condition that block should not slip up,
mω²Rsinθ.cosθ - mgsinθ = µ(mω²R sin²θ + mg cosθ)
mω²R.sinθ.(cosθ-µ.sinθ) =mg(sinθ+µcosθ)
ω²Rsinθ(cosθ - µsinθ) = g(sinθ + µcosθ)
ω = √[g(sinθ + µcosθ)/Rsinθ (cosθ - µ sinθ)]
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