In the figure, a small 0.458 kg block slides down a frictionless surface through height h = 1.05 m and then sticks to a uniform vertical rod of mass M = 0.458 kg and length d = 2.70 m. The rod pivots about point O through angle θ before momentarily stopping. Find θ.
To find the angle θ, we need to consider the conservation of mechanical energy of the system.
The initial mechanical energy of the block is given by the potential energy when it is at height h:
Ep = mgh
where m is the mass of the block, g is the acceleration due to gravity, and h is the height.
The final mechanical energy of the system is the potential energy due to the rotation of the rod, given by:
Er = Iθ
where I is the moment of inertia of the rod and θ is the angle through which it rotates.
Since the block sticks to the rod, the energy loss is accounted for by the increase in potential energy of the combined block and rod system.
Assuming no other external forces act on the system, the mechanical energy is conserved, so:
Ep = Er
Therefore, we can equate the expressions for potential energy:
mgh = Iθ
The moment of inertia of a uniform rod rotating about one end is given by:
I = (1/3)Ml^2
where M is the mass of the rod and l is its length.
Substituting the values into the equation:
mgh = (1/3)Ml^2θ
Now we can solve for θ:
θ = (3mgh)/(Ml^2)
Plugging in the given values:
θ = (3 * 0.458 * 9.8 * 1.05) / (0.458 * 2.70^2)
Simplifying the equation:
θ ≈ 0.922 radians
Therefore, the angle θ through which the rod rotates before momentarily stopping is approximately 0.922 radians.