A simple pendlum bob of mass200g is pulled sideways a horizontal distance of 30cm to a point A. If the length of pendlum is 50cm,calc( i) The p.e of the bob in this position(ii) max speed of the bob when released from that position.
Calc the velocity of oil drop of radius 3.0×10power -6m falling through air of coeffient of viscocity 1.8×10power-5 N.s/m squar,given that the density of oil is800kg/m3 and that of density of air may neglected
A simple pendlum bob of mass200g is pulled sideways a horizontal distance of 30cm to a point A. If the length of pendlum is 50cm,calc( i) The p.e of the bob in this position(ii) max speed of the bob when released from that position.
Solution
(1) Mass of pendulum bob=200Kg
Height bob when raised is=30Cm
Length of the pendulum is=50Cm
P.E as a potential energy
POTENTIAL ENERGY=MASS*GRAVITY*HEIGHT
P.E=MGH,
G as the gravity is 9.8N/Kg, Mass 200Kg, Height is 50Cm, FINISH, change cm into m.
A simple pendlum bob of mass200g is pulled sideways a horizontal distance of 30cm to a point A. If the length of pendlum is 50cm,calc( i) The p.e of the bob in this position(ii) max speed of the bob when released from that position.
Height =50-50cos 37.8=10.4 cm
P.E=mgh=0.2x10x0.104=0.21N Ans
(ii)Conserve energy
Pe ata top =KE at the bottom
Therefore v=square root of 2gh=1.4 m/s Ans