Math Problem Statement

Si el rodillo R de la biela inclinada se mueve a la izquierda a una velocidad constante de VR = 1,2 m/s, determine la velocidad angular del eslabón y la velocidad del rodillo L cuando el ángulo θ es igual a 20°.

Solution

To solve this problem, we need to determine:

  1. The angular velocity of the link ω\omega (between points L and R).
  2. The velocity of point LL.

We know the following information:

  • The velocity of point RR is given as VR=1.2m/sV_R = 1.2 \, \text{m/s} to the left.
  • The length of the link LR=2.3mLR = 2.3 \, \text{m}.
  • The angle θ=20\theta = 20^\circ between the link and the horizontal.

Step 1: Angular Velocity of the Link

The velocity at point RR is related to the angular velocity ω\omega of the link. Using the relative velocity equation for rigid body rotation:

VR=VL+ω×LRV_R = V_L + \omega \times LR

Since VLV_L is the velocity of point LL and we are not given it explicitly, we will decompose the motion:

For point RR, the velocity VRV_R is purely horizontal. For point LL, its velocity will have both horizontal and vertical components. However, the rotational relationship gives us the connection between the angular velocity ω\omega and the linear velocity at RR:

VR=ωLRsin(θ)V_R = \omega \cdot LR \cdot \sin(\theta)

We can rearrange this to solve for the angular velocity ω\omega:

ω=VRLRsin(θ)\omega = \frac{V_R}{LR \cdot \sin(\theta)}

Substitute the values:

ω=1.2m/s2.3msin(20)\omega = \frac{1.2 \, \text{m/s}}{2.3 \, \text{m} \cdot \sin(20^\circ)}

Now calculate ω\omega:

ω=1.22.30.3420=1.20.78661.53rad/s\omega = \frac{1.2}{2.3 \cdot 0.3420} = \frac{1.2}{0.7866} \approx 1.53 \, \text{rad/s}

Step 2: Velocity of Point LL

Once we know ω\omega, we can find the velocity of point LL using the rotational relationship. The velocity at point LL will be:

VL=ωLRV_L = \omega \cdot LR

Substituting the values:

VL=1.53rad/s2.3m3.52m/sV_L = 1.53 \, \text{rad/s} \cdot 2.3 \, \text{m} \approx 3.52 \, \text{m/s}

Thus:

  • The angular velocity ω\omega is approximately 1.53rad/s1.53 \, \text{rad/s}.
  • The velocity of point LL is approximately 3.52m/s3.52 \, \text{m/s}.

Would you like a detailed breakdown of any of these steps or further clarification?


5 related questions:

  1. How would the angular velocity change if the velocity of point RR were doubled?
  2. What would happen to the velocity of point LL if the angle θ\theta increased?
  3. How can we calculate the components of the velocity of point LL in the x and y directions?
  4. Can the link between points LL and RR experience angular acceleration in this scenario?
  5. How is the direction of velocity related to the angular velocity in rigid body motion?

Tip: Always ensure that your angle units match your trigonometric functions—radians or degrees—as this can lead to calculation errors.

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Math Problem Analysis

Mathematical Concepts

Angular velocity
Linear velocity
Trigonometry

Formulas

ω = V_R / (L_R * sin(θ))
V_L = ω * L_R

Theorems

Rigid body rotation kinematics

Suitable Grade Level

Grades 11-12 (High School Physics or Mechanics)