# Rotational Mechanics question with spring

• Ujjwal Basumatary
In summary, the maximum elongation of the spring is 2Mg sin θ, where θ is the angle between the x-axis and the inclined plane.
Ujjwal Basumatary

## Homework Statement

A uniform cylinder of mass ##M## and radius ##R## is released from rest on a rough inclined surface of inclined surface of inclination ##\theta## with the horizontal as shown in the figure. As the cylinder rolls down the inclined surface, what is the maximum elongation in the spring of stiffness ##k##?

2. The attempt at the solution
Here is what I have thought about the problem:

We choose the origin to coincide with the equilibrium position of the spring, the x-axis along the incline and the y-axis perpendicular to it.

Let ##a## be the linear acceleration at an arbitrary instant of time. Also let ##\alpha## be the angular acceleration of the block at the same. Since there is no slipping, we have ##a=R\alpha##.

The equation for tangential motion is, therefore $$Mg\sin\theta - kx-f=Ma$$

For angular motion, $$fR=I\alpha$$

Where,
##x =## displacement from the origin
##g =## acceleration due to gravity
##f =## force of friction
##I =## moment of inertia of the cylinder through its central axis

Substituting the value of ##f## from the second equation into the first, we obtain $$mg\sin\theta - kx- \frac{I\alpha}{R} = Ma$$

But at the extreme position, linear as well as tangential acceleration is zero. Substituting, ##a=\alpha=0## in the above we get $$x=\frac{Mg\sin\theta}{k}$$

But this is the elongation from the origin, and the spring was initially at ##-x##. Therefore maximum elongation is given by$$x_{max} = \frac{2Mg\sin\theta}{k}$$

Am I correct in my reasoning? The answer is given to be this but I suspect I might have missed something in the middle as I am new to rotational mechanics. Any help would be appreciated. Thanks a lot.

#### Attachments

• Untitled.png
4.1 KB · Views: 753
It is a bit unclear to me what you mean by the spring's equilibrium position. You could mean one of two things:
• Where the system is at equilibrium.
• Where the spring is unstretched.
Your argumentation in the end seems to suggest the first, but your argumentation for the location of the zero-acceleration point seems to suggest the latter.

That being said, you could also just do an energy argument, which will avoid any rotational involvement altogether.

Orodruin said:
It is a bit unclear to me what you mean by the spring's equilibrium position. You could mean one of two things:
• Where the system is at equilibrium.
• Where the spring is unstretched.
Your argumentation in the end seems to suggest the first, but your argumentation for the location of the zero-acceleration point seems to suggest the latter.

That being said, you could also just do an energy argument, which will avoid any rotational involvement altogether.
I'm sorry for being unclear. I meant the position where the spring is unstretched.

Then that is where the contraption was released so
Ujjwal Basumatary said:
But this is the elongation from the origin, and the spring was initially at −x.
is not true. The spring was initially at ##x = 0##. However, the point you have computed is not that where the system changes direction, it is where it has zero acceleration.

Orodruin said:
Then that is where the contraption was released so

is not true. The spring was initially at ##x = 0##. However, the point you have computed is not that where the system changes direction, it is where it has zero acceleration.
I see that flaw. Thanks a lot for pointing it out. I found a better and neater solution.

The decrease in gravitational potential energy is equal to the increase in spring potential energy and hence we have $$mgx\sin\theta = \frac{1}{2}kx^2$$
Rearranging yields the answer. I hope this is correct :)

## What is rotational mechanics?

Rotational mechanics is a branch of physics that deals with the motion of objects that rotate or spin around a fixed axis. It involves the study of the forces and torques acting on rotating objects, as well as the angular velocity and acceleration of these objects.

## How does a spring affect rotational motion?

A spring can affect rotational motion by exerting a torque on the rotating object. When a spring is attached to a rotating object, such as a wheel, and compressed or stretched, it generates a restoring torque that causes the object to rotate in the opposite direction. This is known as the spring torque or torsional force.

## What is the relationship between spring stiffness and rotational motion?

The stiffness of a spring, also known as its spring constant, is directly proportional to the amount of torque it produces. This means that the stiffer the spring, the greater the torque it can generate, and the faster the object will rotate. This relationship is described by Hooke's Law, which states that the force exerted by a spring is directly proportional to its displacement.

## How does rotational inertia affect the behavior of a rotating object with a spring?

Rotational inertia, also known as moment of inertia, is a measure of an object's resistance to changes in rotational motion. It depends on the mass and distribution of mass of the object. An object with a larger rotational inertia will require more torque to change its rotational speed. In the case of a rotating object with a spring, the rotational inertia of the object will affect how much the spring torque can accelerate or decelerate the object.

## What are some real-world applications of rotational mechanics with a spring?

Rotational mechanics with a spring has many practical applications, such as in the design of torsion springs for use in mechanical systems, such as car suspensions and door hinges. It is also important in the study of rotational motion in sports, such as in the spin of a golf ball or the rotation of a figure skater. Additionally, understanding rotational mechanics with a spring is crucial in the design and operation of machines, such as engines and turbines, that involve rotating components.

• Introductory Physics Homework Help
Replies
2
Views
686
• Introductory Physics Homework Help
Replies
11
Views
151
• Introductory Physics Homework Help
Replies
3
Views
499
• Introductory Physics Homework Help
Replies
7
Views
427
• Introductory Physics Homework Help
Replies
8
Views
514
• Introductory Physics Homework Help
Replies
95
Views
4K
• Introductory Physics Homework Help
Replies
25
Views
388
• Introductory Physics Homework Help
Replies
14
Views
360
• Introductory Physics Homework Help
Replies
11
Views
234
• Introductory Physics Homework Help
Replies
13
Views
2K