A flexible shaft consists of a 3 mm diameter steel wire in a flexible hollow tube which imposes a frictional torque of 0.04 N m
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Answer:
Moment of inertia of the system is 289.088 kg.m^2
Explanation:
Given:
Mass of the platform which is a uniform disk = 129 kg
Radius of the disk rotating about vertical axis = 1.61 m
Mass of the person standing on platform = 65.7 kg
Distance from the center of platform = 1.07 m
Mass of the dog on the platform = 27.3 kg
Distance from center of platform = 1.31 m
We have to calculate the moment of inertia.
Formula:
MOI of disk =
Moment of inertia of the person and the dog will be mr^2.
Where m and r are different for both the bodies.
So,
Moment of inertia of the system with respect to the axis yy.
⇒
⇒
⇒
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The moment of inertia of the system is 289.088 kg.m^2
Answer:
At time 10.28 s after A is fired bullet B passes A.
Passing of B occurs at 4108.31 height.
Explanation:
Let h be the height at which this occurs and t be the time after second bullet fires.
Distance traveled by first bullet can be calculated using equation of motion
Here s = h,u = 450m/s a = -g and t = t+3
Substituting
Distance traveled by second bullet
Here s = h,u = 600m/s a = -g and t = t
Substituting
Solving both equations
So at time 10.28 s after A is fired bullet B passes A.
Height at t = 7.28 s
Passing of B occurs at 4108.31 height.