2022-04-21T08:15:21+03:00[Europe/Moscow] af true What kind of force is required to keep an object moving in a circle at a constant speed?, An object moving in a circle at a constant speed is accelerating (true or false), equation for velocity of uniform circular motion, angular velocity equation, velocity equation, Time for one rotation equation, angular displacement equation, speed equation, distance equation for an object in uniforn circular motion, angular acceleration equation, centripetal force equation, What is a radian?, Over the top of a hill , On a roundabout, On a banked track, The big dipper, The very long swing, At maximum height flashcards
Circular motion

Circular motion

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  • What kind of force is required to keep an object moving in a circle at a constant speed?
    A constant centripetal force (a force always applied towards the centre of that circle)
  • An object moving in a circle at a constant speed is accelerating (true or false)
    True The direction is always changing hence the velocity is always changing, where acceleration is defined as the change in velocity over time
  • equation for velocity of uniform circular motion
    v = 2pir/T  T = time for one rotation
  • angular velocity equation
    w = linear velocity/radius = 2pi/T = 2pif = Θ/t as f = 1/T f = frequency of the rotation T = object's time period = time for one rotation w = radians per second
  • velocity equation
    v = wr
  • Time for one rotation equation
    T = 2pi/w
  • angular displacement equation
    Θ = wt = 2pit/T = 2pitf
  • speed equation
    v = 2pir/T = wr
  • distance equation for an object in uniforn circular motion
    s =vt = 2pirt/T = Θr
  • angular acceleration equation
    A = w^2 x r = v^2 /r
  • centripetal force equation
    F = mv^2 / r = mw^2xr
  • What is a radian?
    The angle of a circle sector such that the radius is equal to the arc length
  • Over the top of a hill 
    mg - S = mv^2 / r = centrpetal force The vehicle would loose contact if its speed is equal to or geater than a particular speed, then if this happens then the support for is zero so... mg = mv^2 / r thus the vehicle shoud not exceed v^2 = gr
    mg - S = mv^2 / r = centrpetal force The vehicle would loose contact if its speed is equal to or geater than a particular speed, then if this happens then the support for is zero so... mg = mv^2 / r thus the vehicle shoud not exceed v^2 = gr
  • On a roundabout
    The centripetal force is provided by the sideways friction force between the vehicle's tyres and the road surface friction = centripetal force = mv^2 / r limiting force of friction = mv^2 /r =  μmg maximum speed for no slipping v^2 0 = μgr
    The centripetal force is provided by the sideways friction force between the vehicle's tyres and the road surface friction = centripetal force = mv^2 / r limiting force of friction = mv^2 /r =  μmg maximum speed for no slipping v^2 0 = μgr
  • On a banked track
    On a banked track the speed can be high if the centripetal force is weight so the speed for no sideways friction can be calculated horizontal components = (N1 + N2)sinΘ = centripetal force = mv^2 / r vertical components = (N1 + N2)cosΘ = weight = mg  tanΘ = (N1 + N2)sinΘ / (N1 + N2)cosΘ = mv^2 / mgr = v^2/gr the speed for no sideways friction ... v^2 = grtanΘ as for no sideways friction the horizontal component for the normal force acts as the centripetal force 
    On a banked track the speed can be high if the centripetal force is weight so the speed for no sideways friction can be calculated horizontal components = (N1 + N2)sinΘ = centripetal force = mv^2 / r vertical components = (N1 + N2)cosΘ = weight = mg  tanΘ = (N1 + N2)sinΘ / (N1 + N2)cosΘ = mv^2 / mgr = v^2/gr the speed for no sideways friction ... v^2 = grtanΘ as for no sideways friction the horizontal component for the normal force acts as the centripetal force 
  • The big dipper
    The difference between the support force and the weight acts as a centripetal force S - mg = mv^2 / r S = mg + mv^2 / r So the extra force you experience due to circular motion is mv^2 / r
    The difference between the support force and the weight acts as a centripetal force S - mg = mv^2 / r S = mg + mv^2 / r So the extra force you experience due to circular motion is mv^2 / r
  • The very long swing
    The maximum speed occurs when the swing passes through the lowest point 1/2 x mv^2 = mgh v^2 = 2gh The centripetal force is the difference in the support force and weight S - mg = mv^2 / L
    The maximum speed occurs when the swing passes through the lowest point 1/2 x mv^2 = mgh v^2 = 2gh The centripetal force is the difference in the support force and weight S - mg = mv^2 / L
  • At maximum height
    The reaction and the weight provide the centripetal force mg + R = mv^2 / r(radius) at a particular speed v^2 = gr then R = 0 so there would be no force on the person due to the wheel as R =mv^2 / r - mg 
    The reaction and the weight provide the centripetal force mg + R = mv^2 / r(radius) at a particular speed v^2 = gr then R = 0 so there would be no force on the person due to the wheel as R =mv^2 / r - mg