A sphere of mass 25 kg and radius 0.1 m is hung from the celling of a room with the help of a steel wire. The height of the ceiling from the floor is 5.21 m. When the sphere is hung just like a pendulam, its lower surface touches the floor of the room. What will be the velocity of the sphere at the lowest points of its oscillation? The young's modulus for steel =2 times 10^11 N.m^-2, the initial length of the wire =5 m and the radius of the wire=5 times 10^-4 m.

A sphere of mass 25 kg and radius 0.1 m is hung from the celling of a

1.	A sphere of mass 25 kg and radius 0.1 m is hung from the celling of a room with the help of a steel wire. The height of the ceiling from the floor is 5.21 m. When the sphere is hung just like a pendulam, its lower surface touches the floor of the room. What will be the velocity of the sphere at the lowest points of its oscillation? The young's modulus for steel =2 times 10^11 N.m^-2, the initial length of the wire =5 m and the radius of the wire=5 times 10^-4 m.
Answer» Solution :According to fig.1.15 the elongation of the wire at the LOWEST position of the sphere (diameter=0.2 m), `l=5.21-(5+0.2)=0.01m` If the velocity of the sphere at the lowest POINTS of its oscillation is v, then the tension in the wire. `T-mg=(mv^2)/r` or,`T=mg+(mv^2)/r`.....(1) Here,m=mass of the sphere, r=distance of the centre of gravity of the sphere from the POINT of SUSPENSION =5.21 -0.1=5.11 m. Suppose x=radius of the wire. Then, `Y=(TL)/(pix^2l)or,T=(Ypix^2l)/L` From equation (1), we get `mg+(mv^2)/r=(Ypix^2l)/LOR,v^2/r=(Ypir^2l)/(mL)-g` or,`v^2=(Ypix^2lr)/(mL)-rg` `=((2times10^11)times3.14times(5times10^-4)^2times0.01times5.11)/(25times5)-5.11times9.8` =14.10365 `therefore v=3.76 m.s^-1`