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Physics - Kinematics/Rotational Kinematics

  1. Nov 19, 2011 #1
    Hi guys, new to the forums. Got a couple of questions.

    1. Suppose I'm holding a ball and spinning it above my head via a string. I increase the speed at which it rotates, therefore I supply a tangential acceleration. This means that the total acceleration of the system points in a direction that arches at an angle towards the centre of the circle(i.e. my head). In this case, why does the string not slack, and the ball not fly towards the center?

    My thinking is that the total acceleration of the system is just an expression and does not mean that tangential and radial accelerations need to be related. Just like how the total velocity acting on a particle is the modulus of the x and y directional velocities, and yet we consider the particle's velocities separately in their respective dimensions.

    2. How does an air cushion or a stretchable tarp help firemen to catch falling people?

    My reasoning is this: The person has a momentum upon impact, and it must be set to zero at the end (i.e. he comes stationary) i.e. he must deliver an impulse. If he struck the hard ground, the force exerted by the ground on him by Newton's 3rd Law will be extremely large, since F X T = I (I = dmv, F = dmv/T), and the time of impact is extremely short. This causes him to be killed.

    An air cushion helps to lengthen the time (I guess) in which this impulse is transferred out of him, but how does the air cushion exactly help? Is it safe to say that the average force delivered to him per unit time is decreased since the entire momentum is transferred out over a staggered period of time?

    3. I launch a ball at 20m/s, 30 degree incline with respect to the horizontal, off a building 45m high. What is the velocity of the ball (in y-direction) just before it hits the ground? (Sounds like homework, but it's not. It's a concept check question... forgive me if this doesn't qualify)

    I tried using the formula v^2 = u^2 + 2as, modeled in the y-direction.

    This translated to v^2 = (20cos30)^2 - 2(9.8)(45). This is unsolvable if I used the subtraction sign, but I would get the correct answer if I used the plus sign. Why is this so? Acceleration is supposed to be negative in sign. Hope somebody can clarify for me. Thanks.
  2. jcsd
  3. Nov 19, 2011 #2

    Andrew Mason

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    Welcome to PF!.
    The total acceleration is the vector sum of the centripetal and tangential accelerations. The string tension is needed to provide the centripetal force. If you understand why the string does not slacken due to the centripetal force alone, I am not sure why you think the additional tangential acceleration would cause the string to slacken. It is applied at right angles to the radius/string.

    That's basically it. One has to reduce the force that is applied to the body. Force = time rate of change of momentum: F = dp/dt. The change in momentum is determined by the height of the fall. So one has to increase the time over which the momentum changes in order to decrease the force.

    Use energy and the fact that horizontal velocity does not change. So the change in kinetic energy in the y direction is the change in potential energy. KEx + KEy + PE = constant. KEx does not change so ΔKEy = ΔPE = mgΔy. [itex] KE_y=.5mv_y^2[/itex]

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