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Tanzania [10]
4 years ago
15

You hold a barbell in a horizontal position. The barbell’s center of mass is exactly mid-way between your two hands. A friend wa

lks up and attaches a weight on the end near your left hand. How does the torque on the bar exerted by your left hand compare to the torque on the bar from your right hand if the bar remains horizontal and at rest?

Physics
1 answer:
Harrizon [31]4 years ago
6 0

Explanation:

First consider that each hand works as a fulcrum: a pivot point where the barbell can rotate.

Now consider only the left hand. If the center of mass of the barbell is between hands (in the middle) it is displaced respect the fulcrum, therefore the weight which is pushing the bar downwards becomes a rotational force. The same thing happens to the other hand. Now, if more weight is added to the left hand the center of mass is displaced towards the left hand and depending how much weight is added, the center of mass will change its position and therefore the torque each hand experiences changes.

If the center of mass is still between hands: The torque remains almost the same changing only the magnitudes but not the direction.

If the center of mass is on the hand: there is no torque for the left hand because there is no leaver.

If the center of mass is to the left: now the torque changes direction and both hands need to stop it in the same direction.

(see diagram below)

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Answer:

Δτ = 50 N.m

Explanation:

The torque applied on an object is given by the product of the force applied on it and the perpendicular distance between the force and the axis of rotation of the object. That is:

τ = F r

where,

τ = Torque applied on the object

F = Force applied on it

r = distance from axis of rotation

<u>FOR HANDLE SIDE OF DOOR</u>:

τ₁ = F r₁

where,

τ₁ = Torque applied on the object = ?

F = Force applied on it = 100 N

r₁ = distance from axis of rotation = 1 m

Therefore,

τ₁ = (100 N)(1 m)

τ₁ = 100 N.m

<u></u>

<u>FOR MIDDLE OF DOOR</u>:

τ₂ = F r₂

where,

τ₂ = Torque applied on the object = ?

F = Force applied on it = 100 N

r₂ = distance from axis of rotation = 1 m/2 = 0.5 m

Therefore,

τ₂ = (100 N)(0.5 m)

τ₂ = 50 N.m

Now, the difference between the amount of torque in both cases is:

Δτ = τ₁ - τ₂

Δτ = 100 N.m - 50 N.m

<u>Δτ = 50 N.m</u>

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3 years ago
Explain the term equinox in your own words.
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Explanation: UwU O///O

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A 4.0-kilogram ball moving at 8.0 m/s to the right collides with a 1.0-kilogram ball at rest. After the collision, the 4.0-kilog
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Its about momentum. Momentum (p)=mass(m)xvelocity(v)
So for the first ball P=4x8=32kgm/s
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Momentum has to be conserved
After the collision the momentum of the 4kg ball is 4x4.8=19.2kgm/s
As momentum is conserved 32-19.2=12.8kgm/s remaining
So rearrange for velocity so v=p/m=12.8/1=12.8m/s for the 1kg ball 
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