Q#7
If the road of the previous problem is horizontal (No Banking), what should be the minimum friction coefficient so that a scooter going at 18 km /hr does not skid?
Answer:
In such case minimum frictional force = Centripetal force
µN = mv²/r
µmg=mv²/r (∵ N=mg)
µ = v²/gr = 5²/(10 x 10) = 25/100 = 1/4 = 0.25
Q#8
A circular road of radius 50 m has the angle of banking equal to 30°. At what speed should a vehicle go on this road so that the friction is not used?
Answer :
As in the case of problem 6 where no frictional force is used to turn,
tanθ = v²/gr
v² = grtanθ
here, r = 50 m, θ = 30° & take g = 10 m/s²
it gives v² = 10 x 50 x tan30° = 500 x (1/√3) = 500/1.732 ≈ 269
v = 17 m/s²
So, at 17 m/s² a vehicle would not require friction to turn on this road.
Q#9
In the Bohr model of hydrogen atom, the electron is treated as a particle going in a circle with the center at the proton. The proton itself is assumed to be fixed in an inertial frame. The centripetal force is provided by the Coulomb attraction. In the ground state, the electron goes around the proton in a circle of radius 5.3 x 10-11 m. Find the speed of the electron in the ground state. Mass of the electron = 9.1 x 10-31 kg and charge of the electron = 1.6 x 10-19 C.
Answer :
Here r = 5.3 x 10-11 m
m = 9.1x10-31 kg
Charge on electron or proton, q = 1.6 x 10-19 C v = speed of electron = ?
Force of coloumb attraction on the electron = kq²/r² (constant, k = 9 x 109 m²/C² )
This coloumb attraction force is the centripetal force = mv²/r giving the electron required turn on the circular path, thus
mv²/r = kq²/r²
v² = kq²/mr = 9 x 109 x (1.6 x 10-19)²/[9.1 x 10-31 x 5.3 x 10-11]
= 0.478 x 1013
= 4.78 x 1012
v = 2.186 x 106 m/s
v ≈ 2.2 x 106 m/s
Q#10
A stone is fastened to one end of a string and is whirled in a vertical circle of radius R. Find the minimum speed of the stone can have at the highest point of the circle.
Answer:
At the highest point of the circle minimum speed should be such that its weight is able to turn it on the given circle of radius R.
i.e. mg = mv²/R
v² = gR
v = √(gR)
Q#11
A ceiling fan has a diameter (of the circle through the outer edges of the three blades) of 120 cm and rpm 1500 at full speed. Consider a particle of mass 1 g sticking at the outer end of a blade. How much force does it experience when the fan runs at full speed? Who exerts this force on the particle? How much force does the particle exert on the blade along its surface?
Answer:
Max Angular Speed of the fan
ω = 1500 rpm = 1500 x 2π/(60 rad/s) = 50π rad/s
Radius r = 120/2 cm = 60 cm = 0.60 m
Mass of the particle = 1 g = 0.001 kg
Force experienced by the particle = mω²r
At full speed, this force = 0.001 x (50π)² x 0.60 N
= 0.001 x 2500 x 0.60 π² = 2.5 x 0.60π²
= 1.5π² N = 1.5 x 3.14 x 3.14 N
= 14.8 N
Since the particle is in a circular motion, it has a radial/centripetal acceleration (towards the center). So the force on the particle is also towards the center. Due to the inertia of the particle, it tends to move along the tangential direction with a uniform speed. Since it is stuck to the blade, the blade of the fan exerts a centripetal force on the particle and forces it to move in a circle.
It is the inertia of the particle that exerts the same magnitude of the force on the blade, i.e. 14.8 N is exerted by the particle on the blade along its surface.
Q#12
A mosquito is sitting on an LP record disc rotating on a turntable at 33⅓ revolutions per minutes. The distance of the mosquito from the center of the turntable is 10 cm. Show that the friction coefficient between the record and the mosquito is greater than π²/81. Take g= 10 m/s².
Answer:
Angular speed of the mosquito, ω = 33⅓ rpm = 33⅓ x 2π/60 = 200π/180 = 10π/9 rad/s.
Radius, r = 10 cm = 0.10 m.
Outward force on the mosquito = mω²r (where 'm' is mass of the mosquito).
= m.(100π²/81) x 0.10
= 10π²m/81
Weight of mosquito = mg = 10m = Normal force
So Limiting frictional force = µ.10m
Since the mosquito is sitting on the record, this limiting Frictional force > centrifugal force on the mosquito.
µ(10m) > (10π²m)/81
µ > π²/81.
Hence proved.
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