An inclined plane makes an angle of 60 degrees with horizontal. A disc rolling down this inclined plane without slipping has linear acceleration to

To determine the linear acceleration of a disc rolling down an inclined plane at an angle of 60 degrees, we can apply some fundamental principles of physics, particularly Newton's laws of motion and the concepts of rotational dynamics. Let’s break this down step by step.

Understanding the Forces at Play

When a disc rolls down an inclined plane, two main forces act on it: gravitational force and the normal force. The gravitational force can be decomposed into two components:

• Parallel Component: This is the force that pulls the disc down the slope, calculated as \( mg \sin(\theta) \), where \( m \) is the mass of the disc, \( g \) is the acceleration due to gravity, and \( \theta \) is the angle of the incline.
• Perpendicular Component: This is the force acting perpendicular to the incline, calculated as \( mg \cos(\theta) \).

Applying Newton's Second Law

According to Newton's second law, the net force acting on the disc along the incline can be expressed as:

Net Force = Mass × Acceleration

For the disc, the net force down the incline is the parallel component of the gravitational force:

Net Force = mg \sin(\theta)

Thus, we can write:

mg \sin(\theta) = ma

Here, \( a \) is the linear acceleration of the disc. We can simplify this to find \( a \):

a = g \sin(\theta)

Considering Rotational Motion

Since the disc is rolling without slipping, we also need to consider its rotational motion. The moment of inertia \( I \) for a solid disc about its center is given by:

I = (1/2) m r²

where \( r \) is the radius of the disc. The angular acceleration \( \alpha \) is related to the linear acceleration \( a \) by the equation:

a = r \alpha

Using the torque \( \tau \) caused by the gravitational force, we have:

τ = Iα = r(mg \sin(\theta))

Substituting \( I \) and rearranging gives us:

(1/2) m r² α = r(mg \sin(\theta))

Dividing both sides by \( r \) (assuming \( r \neq 0 \)) leads to:

(1/2) m r α = mg \sin(\theta)

Now substituting \( a = r \alpha \) into this equation gives:

(1/2) m a = mg \sin(\theta)

From this, we can solve for \( a \):

a = (2/3) g \sin(\theta)

Calculating the Acceleration

Now, substituting \( \theta = 60^\circ \) into the equation:

sin(60^\circ) = √3/2

Thus, the linear acceleration \( a \) becomes:

a = (2/3) g (√3/2)

Substituting \( g \approx 9.81 \, \text{m/s}² \):

a ≈ (2/3) × 9.81 × (√3/2) ≈ 5.67 \, \text{m/s}²

Final Thoughts

In summary, the linear acceleration of a disc rolling down an inclined plane at an angle of 60 degrees is approximately 5.67 m/s². This result illustrates the interplay between linear and rotational dynamics, showcasing how both aspects of motion contribute to the overall behavior of rolling objects.