Explain Force & Acceleration for Tension in Rope Around Cylinder Drum

  • Thread starter tnutty
  • Start date
  • Tags
    Explain
In summary, a 55kg mass tied to a massless rope wrapped around a solid cylindrical drum and mounted on a frictionless horizontal axle falls with an acceleration of 1.0m/s^2 when released. To find the tension in the rope, the formula m*g - T is used, where m is the mass, g is the acceleration due to gravity, and T is the tension. In this case, the tension is equal to 484N. The net acceleration is 1m/s/s, which means that the drum is decelerating the mass at 8.8m/s^2. This is because the tension in the rope is m*g - m*a, where a is the net acceleration. Therefore,
  • #1
tnutty
326
1

Homework Statement



A 55kg mass is tied to a massless rope wrapped around a solid cylindrical drum. The drum is mounted on a frictionless horizontal axle. When the mass is released, it falls with acceleration 1.0m/s^2 .

Find the tension in the rope.

my answer : 484N

the way I got is by doing this : 55(9.8-1.0).

I did this because my friend told me this but am confused on
why I have to subtract g from acceleration. Please explain in details.
 
Physics news on Phys.org
  • #2
tnutty said:

Homework Statement



A 55kg mass is tied to a massless rope wrapped around a solid cylindrical drum. The drum is mounted on a frictionless horizontal axle. When the mass is released, it falls with acceleration 1.0m/s^2 .

Find the tension in the rope.

my answer : 484N

the way I got is by doing this : 55(9.8-1.0).

I did this because my friend told me this but am confused on
why I have to subtract g from acceleration. Please explain in details.

If there was no retarding force then the acceleration from gravity would have been unaffected. So for acceleration to be just 1, then there must be a counteracting force that relates to 9.8 - 1 ... in order to end with 1 as your acceleration.
 
  • #3
This statement confuses me :
When the mass is released, it falls with acceleration 1.0m/s^2 .

Its falling down at 1 meters per second per second. This makes me think that
with gravity counted, it falls at 1 m/s/s.
 
  • #4
tnutty said:
This statement confuses me :
When the mass is released, it falls with acceleration 1.0m/s^2 .

Its falling down at 1 meters per second per second. This makes me think that
with gravity counted, it falls at 1 m/s/s.

Right it does mean that.

It means that gravity is accelerating it at 9.8 and the drum is decelerating it at 8.8. The 8.8 then is the Tension in the cable.

Otherwise if it was free falling it would be slack.
 
  • #5
Why should it not be decelerating at 1m/s/s instead of accelerating at 1m/s/s.
It seems like if if its accelerating then we should add 1.0 instead of subtract.
 
  • #6
Ftotal = ma …

and Ftotal = mg - T :wink:
 
  • #7
Ya but why is tension m*8.8 instead of m*10.8. After all it is falling 1.0m/s/s with the
gravity? It just seems counter intuitive to me.
 
  • #8
tnutty said:
Ya but why is tension m*8.8 instead of m*10.8. After all it is falling 1.0m/s/s with the
gravity? It just seems counter intuitive to me.

No. It's not falling 1m/s2 faster than gravity.

It's falling at just 1m/s2

If it was static what would the Tension be? m*g right? And it's not moving.

So as TinyTim pointed out the tension will be m*g less the net acceleration of 1 m/s2
 
  • #9
So the net acceleration is 1m/s/s.

If so then why isn't m*a = m*(1). Isn't a the net acceleration?
 
  • #10
tnutty said:
So the net acceleration is 1m/s/s.

If so then why isn't m*a = m*(1). Isn't a the net acceleration?

Yes …
tiny-tim said:
Ftotal = ma …

and Ftotal = mg - T :wink:

ma = mg - T,

so m*(1) = mg - T, and T = mg - m*(1).

No? :smile:
 

FAQ: Explain Force & Acceleration for Tension in Rope Around Cylinder Drum

What is force in relation to tension in a rope around a cylinder drum?

Force is a physical quantity that describes the push or pull on an object. In the context of tension in a rope around a cylinder drum, force refers to the amount of energy exerted on the rope in order to create tension.

How does the force applied affect the tension in the rope around the cylinder drum?

The force applied to the rope directly affects the tension in the rope. As the force increases, the tension in the rope also increases. Similarly, as the force decreases, the tension in the rope decreases.

What is acceleration and how does it relate to tension in a rope around a cylinder drum?

Acceleration is the rate of change of an object's velocity. In the context of tension in a rope around a cylinder drum, acceleration is the change in speed of the rope as a result of the force applied. This change in speed directly affects the tension in the rope.

Can the tension in the rope around the cylinder drum change without a change in force or acceleration?

No, the tension in the rope cannot change without a change in force or acceleration. This is because the tension in the rope is directly related to the force applied and the acceleration of the rope.

How does the diameter of the cylinder drum affect the tension in the rope?

The diameter of the cylinder drum does not directly affect the tension in the rope. However, it can indirectly affect the tension by changing the amount of force needed to create tension. A larger diameter drum may require a greater force to create the same amount of tension as a smaller diameter drum.

Similar threads

Back
Top