Lessons 18–19 · 2 lessons · N. Sh. Turdiyev, K. A. Tursunmetov, A. G. Ganiyev, K. T. Suyarov, J. E. Usarov, A. K. Avliyoqulov. Physics Grade 10, 1st edition. Niso Poligraf Publishing House, Tashkent, 2017
19
Dynamics of rotational motion
Textbook: pp. 65–67
GoalCalculate the dynamics of circular motion: centripetal force, banked roads and speed in a vertical circle.
New words
centripetal force · markazga intilma kuchcentrifugal inertial force · markazdan qochma inersiya kuchispeed limit on a turn · tezlik chegarasivertical circle · vertikal aylana
Explanation
In uniform motion along a circle the net force on a body is directed along the radius toward the centre: Fc = mυ²/R = mω²R = m·4π²R/T². It is not a separate kind of force but the resultant of the actual forces: string tension, friction, and components of weight and reaction. In a rotating (non-inertial) frame an equal and opposite centrifugal inertial force Fcf = mυ²/R is introduced. On a horizontal turn friction holds the car: μmg ≥ mυ²/R, so the greatest speed is υ = √(μgR). If the road is banked inward at an angle α, the speed at which no friction is needed is υ = √(gR·tanα). In a vertical circle a body on a string or in a bucket will not fall at the top if mg ≤ mυ²/R, i.e. υ ≥ √(gR); that is why water does not spill from a swung bucket. Drivers must slow down on turns on wet roads or ice, where μ is small.
Worked examples
A car (1000 kg) takes a turn of R = 50 m at 10 m/s: Fc = 1000·100/50 = 2000 N. On a wet road with μ = 0.2 the greatest speed is √(0.2·10·50) = √100 = 10 m/s.
A bucket (R = 0.9 m) is swung in a vertical circle: for the water not to fall at the top υ ≥ √(gR) = √(10·0.9) = 3 m/s. For a turn of radius R = 160 m and speed 20 m/s, the friction-free banking angle is given by tanα = υ²/(gR) = 400/1600 = 0.25, α ≈ 14°.
Class activity
In the school yard, only under the teacher’s supervision: put a little water in a small plastic cup tied to a string and swing it slowly in a vertical circle. Feel the smallest speed at which the water does not spill. Keep everyone clear and do not swing fast.
Practice
1
What is the formula of the centripetal force and where does it point?
Fc = mυ²/R = mω²R; toward the centre of the circle.
2
A 0.5 kg stone moves on a string in a horizontal circle of R = 1 m at 4 m/s. What is the tension (N)?
8
3
A turn with R = 100 m on a wet road (μ = 0.1, g = 10 m/s²). What is the greatest speed (m/s)? (υ = √(μgR))
10
4
Why is the outer rail raised above the inner one on railway curves?
The horizontal component of the reaction force supplies the centripetal force, so the wheel flange does not press hard on the rail.