A wheel of radius R and mass M is rolling without slipping on a horizontal surface. If it has a linear speed v, what is its total kinetic energy? (2022)
Practice Questions
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Q1
A wheel of radius R and mass M is rolling without slipping on a horizontal surface. If it has a linear speed v, what is its total kinetic energy? (2022)
(1/2)Mv²
(1/2)Mv² + (1/2)(Iω²)
(1/2)Mv² + (1/2)(Mv²)
(1/2)Mv² + (1/2)(Mv²/2)
The total kinetic energy is the sum of translational and rotational kinetic energy. K.E. = (1/2)Mv² + (1/2)(Iω²) where I = (1/2)MR² for a solid cylinder.
Questions & Step-by-step Solutions
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Q
Q: A wheel of radius R and mass M is rolling without slipping on a horizontal surface. If it has a linear speed v, what is its total kinetic energy? (2022)
Solution: The total kinetic energy is the sum of translational and rotational kinetic energy. K.E. = (1/2)Mv² + (1/2)(Iω²) where I = (1/2)MR² for a solid cylinder.
Steps: 9
Step 1: Understand that the total kinetic energy of the wheel is made up of two parts: translational kinetic energy and rotational kinetic energy.
Step 2: The formula for translational kinetic energy (K.E.) is (1/2)Mv², where M is the mass of the wheel and v is its linear speed.
Step 3: The formula for rotational kinetic energy is (1/2)(Iω²), where I is the moment of inertia and ω is the angular speed.
Step 4: For a solid cylinder (which is the shape of the wheel), the moment of inertia I is (1/2)MR², where R is the radius of the wheel.
Step 5: Since the wheel is rolling without slipping, the relationship between linear speed v and angular speed ω is given by ω = v/R.
Step 6: Substitute ω = v/R into the rotational kinetic energy formula: (1/2)(Iω²) becomes (1/2)((1/2)MR²)(v/R)².
Step 7: Simplify the rotational kinetic energy: (1/2)((1/2)MR²)(v²/R²) = (1/4)Mv².
Step 8: Now, add the translational and rotational kinetic energies together: Total K.E. = (1/2)Mv² + (1/4)Mv².
Step 9: Combine the terms: Total K.E. = (2/4)Mv² + (1/4)Mv² = (3/4)Mv².
