A simple pendulum consists of bob of mass 0.085 kg attached tostrong of length 1

Question

A simple pendulum consists of bob of mass 0.085 kg attached tostrong of length 1.5 m. Pendulum is raised to point Q locatedon right and which is .08 m above its lowest position. Thependulum is released so that it oscillates with small amplitude between points P on left and Q on right. Calculate speed v of bob at its lowest position Calculate tension in string when bob is passing through itslowest position Describe one modification that could be made to double periodof oscillation What forces act on bob at point P What forces act on bob when it is in motion at its lowestposition? A simple pendulum consists of bob of mass 0.085 kg attached tostrong of length 1.5 m. Pendulum is raised to point Q locatedon right and which is .08 m above its lowest position. Thependulum is released so that it oscillates with small amplitude between points P on left and Q on right. Calculate speed v of bob at its lowest position Calculate tension in string when bob is passing through itslowest position Describe one modification that could be made to double periodof oscillation What forces act on bob at point P What forces act on bob when it is in motion at its lowestposition?

Explanation / Answer

   a.   The kinetic energy ofbob at lowest point is equal to potential energy at highest point (law of conservation of energy).    =>   m * g *h   =   (1/2) * m *v2    m   =   massofbob   =   0.085   kg,   h   =   heightof bob above lowestpoition   =   0.08 m    =>   v   =   (2* g * h)   =   ( 2 * 9.8 *0.08)   =   1.25   m/s    b.   Tension in string atlowestpoint   T   =   Weightof bob   +   centripetal force                                                                  =   m* g   +   (1/2) * m * v2 /L                                                                   =   0.085* ( 9.8 + 0.5 * 1.252 /1.5)   =   0.877   N    c.   Timeperiod   T   =   2(L/g),      L   =   Lengthof pendulum          Given   T'   =   2T          =>   T'/T   =   2   =   2(L'/g)/2(L/g)      =>   L'   =   4* L       as only L is variable thatcan be modified or changed.       Hence to double the timeperiod of a simple pendulum its lenght must be made fourtimes.    d.   At extreme, the bobexperiences forces of weight and tension, since there is nocentripetal force as the velocity of bob is zero there.    e.   Centrifugal force andweight.                                                                   =   0.085* ( 9.8 + 0.5 * 1.252 /1.5)   =   0.877   N    c.   Timeperiod   T   =   2(L/g),      L   =   Lengthof pendulum          Given   T'   =   2T          =>   T'/T   =   2   =   2(L'/g)/2(L/g)      =>   L'   =   4* L       as only L is variable thatcan be modified or changed.       Hence to double the timeperiod of a simple pendulum its lenght must be made fourtimes.    d.   At extreme, the bobexperiences forces of weight and tension, since there is nocentripetal force as the velocity of bob is zero there.    e.   Centrifugal force andweight.

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