A Canadian college student who has taken an astronomy class goes home for the holidays and persuades his parents to let him borr
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Answer:
the maximum is I₁ axis of rotation at the end
the minimum moment is I₂ axis of rotation at the center of mass
Explanation:
For this exercise we use the definition moment of inertia
I = ∫ r² dm
for bodies of high symmetry it is tabulated; In this case we can approximate a broomstick to a thin rod, the moment of inertia with respect to a perpendicular axis when varying are
at one end
I₁ = ⅓ mL²
in in center
I₂ = m L²
There is another possible axis of rotation around the axis of the broom, in this case we have a solid cylinder
I₃ = m r²
remember that the diameter of the broom is much smaller than its length, therefore this moment of inertia is very small
when examining the different moments of inertia:
the maximum is I₁ axis of rotation at the end
the minimum moment is I₂ axis of rotation at the center of mass
The acceleration is found to be 2 m/s², final velocity after 10 seconds is 20 m/s and the final position after 10 seconds is 100 m away from the starting point.
Answer:
Explanation:
As per Newton's second law of motion, acceleration of any object is directly proportional to the net external unbalanced force acting on that object and inversely proportional to the mass of the object.
Since there are two forces acting on the box in opposite direction, the net force will be the difference of horizontal and frictional force acting on the object.
Net force = Horizontal force - Frictional force = 18 N - 8 N = 10 N.
Now, from second law of motion,
So, acceleration = 10 N /5 kg = 2 m/s².
Since, acceleration exerted by the box is found to be 2 m/s², we can determine the final velocity of the object after 10 seconds using the first equation of motion.
v = u + at, Here v is the final velocity and u is the initial velocity which is zero for the present case. Other parameters like a is found to be 2 m/s² and time is 10 seconds.
So Final velocity v = 0+(2×10)=20 m/s.
And the final position can be determined using the second equation of motion.
s = ut+1/2at²
Final position = (0×10)+(0.5×2×10×10)= 100 m.
So the final position is 100 m.
Thus, the acceleration is found to be 2 m/s², final velocity after 10 seconds is 20 m/s and the final position after 10 seconds is 100 m away from the starting point.