a uniform rod of length 1.5m is placed over a wedge at 0.5m from one end .a force of 100 N is applied at its one end near the we
1 answer:
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Answer:
2.85 rad/s
Explanation:
5 cm = 0.05 m
20 g = 0.02 kg
When dropping the 2nd object at a distance of 0.05 m from the center of mass, its corrected moments of inertia is:
So the total moment of inertia of the system of 2 objects after the drop is:
From here we can apply the law of angular momentum conservation to calculate the post angular speed
Answer:
570 N
Explanation:
Draw a free body diagram on the rider. There are three forces: tension force 15° below the horizontal, drag force 30° above the horizontal, and weight downwards.
The rider is moving at constant speed, so acceleration is 0.
Sum of the forces in the x direction:
∑F = ma
F cos 30° - T cos 15° = 0
F = T cos 15° / cos 30°
Sum of the forces in the y direction:
∑F = ma
F sin 30° - W - T sin 15° = 0
W = F sin 30° - T sin 15°
Substituting:
W = (T cos 15° / cos 30°) sin 30° - T sin 15°
W = T cos 15° tan 30° - T sin 15°
W = T (cos 15° tan 30° - sin 15°)
Given T = 1900 N:
W = 1900 (cos 15° tan 30° - sin 15°)
W = 570 N
The rider weighs 570 N (which is about the same as 130 lb).
