Flag Modern Physics> 4 A slotted link is rotating about fixed ...
question mark

4 A slotted link is rotating about fixed pivot O with a counterclockwise angular velocity of 3 rad/s, and a clockwise angular acceleration of 2 rad/s2. The movement of the link is causing a rod to slide along the curved channel, as shown. The radius of the channel as a function of ? is given by, R = 0.7? (with R in meters and ? in radians). Determine the radial and circumferential velocity and acceleration components of the rod at ? = 45° .

Vicki Sharma , 12 Years ago
Grade 12
anser 1 Answers
Askiitians Tutor Team

To solve this problem, we need to break down the motion of the slotted link and how it affects the rod sliding along the curved channel. We will calculate the radial and circumferential components of both velocity and acceleration at the specified angle of 45 degrees (which we will convert to radians for our calculations).

Understanding the Geometry and Motion

The radius of the channel as a function of the angle θ is given by:

R(θ) = 0.7θ

At θ = 45 degrees, we first convert this angle to radians:

θ = 45° × (π/180) = π/4 radians

Now, substituting θ into the radius function:

R(π/4) = 0.7 × (π/4) ≈ 0.549 m

Calculating the Velocity Components

The angular velocity (ω) is given as 3 rad/s, and the radial distance (R) at θ = π/4 is approximately 0.549 m. The velocity components can be derived as follows:

  • Circumferential Velocity (Vθ): This is given by the formula:

Vθ = R × ω

Substituting the values:

Vθ = 0.549 m × 3 rad/s ≈ 1.647 m/s

  • Radial Velocity (Vr): Since the radius is changing with time, we need to consider the time derivative of R:

To find the radial velocity, we first need to calculate the derivative of R with respect to time:

R = 0.7θ

Taking the time derivative:

Vr = dR/dt = d(0.7θ)/dt = 0.7(dθ/dt) = 0.7ω

Substituting the angular velocity:

Vr = 0.7 × 3 rad/s = 2.1 m/s

Determining the Acceleration Components

Next, we will calculate the radial and circumferential acceleration components. The angular acceleration (α) is given as 2 rad/s².

  • Circumferential Acceleration (Aθ): This is given by:

Aθ = R × α

Substituting the values:

Aθ = 0.549 m × 2 rad/s² ≈ 1.098 m/s²

  • Radial Acceleration (Ar): This is composed of two parts: the centripetal acceleration and the tangential acceleration:

Ar = (Vθ²/R) + (R × α)

Calculating the centripetal acceleration:

Vθ²/R = (1.647 m/s)² / 0.549 m ≈ 4.706 m/s²

Now, calculating the tangential acceleration:

R × α = 0.549 m × 2 rad/s² ≈ 1.098 m/s²

Thus, the total radial acceleration is:

Ar = 4.706 m/s² + 1.098 m/s² ≈ 5.804 m/s²

Summary of Results

At θ = 45° (or π/4 radians), we find:

  • Circumferential Velocity (Vθ): ≈ 1.647 m/s
  • Radial Velocity (Vr): ≈ 2.1 m/s
  • Circumferential Acceleration (Aθ): ≈ 1.098 m/s²
  • Radial Acceleration (Ar): ≈ 5.804 m/s²

This breakdown provides a comprehensive understanding of the motion of the rod as it slides along the curved channel, influenced by the angular velocity and acceleration of the slotted link.

ApprovedApproved
Last Activity: 10 Months ago
star
LIVE ONLINE CLASSES

Prepraring for the competition made easy just by live online class.

tv

Full Live Access

material

Study Material

removal

Live Doubts Solving

assignment

Daily Class Assignments