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Brums [2.3K]
3 years ago
7

Four uniform objects having the same mass and diameter are released simultaneously from rest at the same distance above the bott

om of a hill and roll down without slipping. the objects are a solid sphere, a solid cylinder, a hollow cylinder and a thin-walled hollow cylinder. which of these objects will be the last one to reach the bottom of the hill?
Physics
1 answer:
Jet001 [13]3 years ago
8 0
The one with the majority of its mass furthest away from its center will reach the bottom last. The thin walled cylinder has all of its mass at the edge of itself, meaning its mass is the furthest from its center. If we take a look at the solid cylinder, it has some mass at its edge, but it also has a lot of mass very close to its center so it will roll faster than the thin walled cylinder. And the thin walled cylinder will be slower than the normal hollow cylinder because even though the hole cylinder doesn't have any mass at its center it does have some mass closer to its center than the thin walled cylinder.

So the last one to reach the bottom is the thin walled cylinder.
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A one-piece cylinder has a core section protruding from the larger drum and is free to rotate around its central axis. A rope wr
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Answer:

Magnitude the net torque about its axis of rotation is 2.41 Nm

Solution:

As per the question:

The radius of the wrapped rope around the drum, r = 1.33 m

Force applied to the right side of the drum, F = 4.35 N

The radius of the rope wrapped around the core, r' = 0.51 m

Force on the cylinder in the downward direction, F' = 6.62 N

Now, the magnitude of the net torque is given by:

\tau_{net} = \tau + \tau'

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\tau = Torque due to Force, F

\tau' = Torque due to Force, F'

tau = F\times r

tau' = F'\times r'

Now,

\tau_{net} = - F\times r + F'\times r'

\tau_{net} = - 4.35\times 1.33 + 6.62\times 0.51 = - 2.41\ Nm

The net torque comes out to be negative, this shows that rotation of cylinder is in the clockwise direction from its stationary position.

Now, the magnitude of the net torque:

|\tau_{net}| = 2.41\ Nm

 

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