A dumbbell-shaped object is composed by two equal masses, m, connected by a rod of negligible mass and length r. If I_1 is the m
1 answer:
Answer:
I2>I1
Explanation:
This problem can be solved by using the parallel axis theorem. If the axis of rotation of a rigid body (with moment of inertia I1 at its center of mass) is changed, then, the new moment of inertia is gven by:
where M is the mass of the object and d is the distance of the new axis to the axis of the center of mass.
It is clear that I2 is greater than I1 by the contribution of the term Md^2.
I2>I1
hope this helps!!
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AS
work done =W = F.d = F d cosФ (Ф is angle between force F and displacement d) If a body/object is moving on a smooth surface (friction-less surface ) .There is no force acting on that body. F=0 so W=FdcosФ= (0)dcosФ ⇒ W=0
Now if a body is facing some amount of force but under the action of force there is no displacement covered. d=0 so W =FdcosФ= F(0)cosФ ⇒W=0
example: A person is applying a force on rigid wall but wall remains at rest there is no displacement occurs in wall.
The third term upon which work done dependent is angle between force and displacement i.e Ф. If Ф=90° then W= FdcosФ= Fdcos90⇒ W=0 ( as cos 90°=0)
