Question

Suppose the smaller pulley of the previous problem has its radius 5⋅0 cm and moment of inertia 0⋅10 kg-m2. Find the tension in the part of the string joining the pulleys.

Answer 1

Answer:

tension equals to 400 Newton

Explanation:

given :r=5×10^-2m ,I=0.1kg/m2

tension ko mg balance karega if smaller pulley are not moving

T=mg. then first find m

for smaller pulley moment of inertia=mr^2

0.1=m(5×10^-2)^2

0.1=m25×10^-4

m=40

T=mg=40×10=400N

hope Karunga ki ans right hoga ek bar ans key me check kar Lena