Here's the question you clicked on:
bii17
A bullet emerges from a 9mm pistol caliber spinning on its axis. Explain how this prevents from tumbling and keeps the streamlined end pointed forward..
:)) here's another question.. The work done by a force is the product of force and distance. Does this means that the torque and work are equivalent?
One reason for that - GIVING GYROSCOPIC STABILITY to the bullet . This stability is due to angular momentum. When a body is spinning around some axis, we say the angular momentum is pointing along this axis with a magnitude equal to the magnitude of the angular momentum. When a body has angular momentum is some direction, much like with regular momentum, it doesn't like to have its magnitude or direction changed. One particular note to this is precession, where the angular momentum of a body spins about a 3rd axis, but it doesn't like to have this 3rd axis changed either. So, when the bullet has some spin, it has some angular momentum in the direction of its motion. This spinning adds stability, because the bullet itself doesn't want to turn on some other axis, thus changing the direction of its angular momentum, so it stays pointing straight. Since it stays pointing straight, it is more aerodynamic, so it flies for longer.
Answer to SECOND q (btw whre is the medl for the 1-st...) is as follows Torque equals VECTOR-PRODUCT of FORCE-times-Distance-from center WORK equals SCALAR product of FORCE-time-DITANCE. It means that :
Torque increases WIT the angle, because SIn alpha INCRESES with the angle. In contrast- WORK DECREASES with the angle because Cos alpha decreases with the angle
@bii17 I have answered both of your questions.