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15.13: Equation of Motion: General Plane motion - Problem Solving

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Equation of Motion: General Plane motion - Problem Solving
 
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15.13: Equation of Motion: General Plane motion - Problem Solving

Consider a lawn roller with a mass of 100 kg, a radius of 0.2 meters, and a radius of gyration of 0.15 meters. A force of 200 N is applied to this roller, angled at 60 degrees from the horizontal plane. What will be the angular acceleration of the lawn roller?

The friction between the roller and the ground is characterized by two coefficients. The static friction coefficient is 0.15, while the kinetic friction coefficient is 0.1. These values are crucial in understanding the interaction between the roller and the surface on which it moves. The concept of rolling without slipping is assumed in this scenario. In this context, the moment of the point of zero instantaneous velocity is calculated. This calculation is made using the horizontal component of the applied force.

Equation 1

The point of zero instantaneous velocity also serves as a reference for the calculation of the moment of inertia, derived using the parallel axis theorem.

Equation 2

Once the moment of inertia is determined, it is substituted into the moment equation, subsequently providing the sought-after value of angular acceleration.

Equation 2

However, it is important to note that the assumption of rolling without slipping is only valid under certain conditions. Specifically, the frictional force resulting from the movement of the lawn roller's center must be lower than the maximum static frictional force. Only then can the roller move without slipping.

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Angular Acceleration Of The Lawn Roller: Given: - Mass Of The Lawn Roller M = 100 Kg - Radius Of The Lawn Roller R = 0.2 M - Radius Of Gyration K = 0.15 M - Applied Force F = 200 N - Angle Of The Applied Force θ = 60° - Static Friction Coefficient μs = 0.15 - Kinetic Friction Coefficient μk = 0.1 Step 1: Calculate The Moment Of The Applied Force About The Center Of The Roller. Moment Of The Applied Force = F × R × Sin(θ) Moment Of The Applied Force = 200 N × 0.2 M × Sin(60°) = 34.64 N·m Step 2: Calculate The Moment Of Inertia Of The Roller About Its Center Of Mass. Moment Of Inertia I = M × K² Moment Of Inertia I = 100 Kg × (0.15 M)² = 2.25 Kg·m² Step 3: Calculate The Angular Acceleration Of The Roller. Angular Acceleration α = Moment Of The Applied Force / Moment Of Inertia Angular Acceleration α = 34.64 N·m / 2.25 Kg·m² = 15.4 Rad/s² Therefore The Angular Acceleration Of The Lawn Roller Is 15.4 Rad/s²

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