If a rolling object has a radius of R and rolls with a speed v, what is its kinetic energy?
Practice Questions
1 question
Q1
If a rolling object has a radius of R and rolls with a speed v, what is its kinetic energy?
(1/2)mv^2
(1/2)mv^2 + (1/2)Iω^2
(1/2)mv^2 + (1/2)(1/2)mR^2(v/R)^2
None of the above
The total kinetic energy of a rolling object is the sum of translational and rotational kinetic energy, which simplifies to (1/2)mv^2 + (1/4)mv^2 = (3/4)mv^2.
Questions & Step-by-step Solutions
1 item
Q
Q: If a rolling object has a radius of R and rolls with a speed v, what is its kinetic energy?
Solution: The total kinetic energy of a rolling object is the sum of translational and rotational kinetic energy, which simplifies to (1/2)mv^2 + (1/4)mv^2 = (3/4)mv^2.
Steps: 8
Step 1: Understand that a rolling object has two types of motion: it moves forward (translational motion) and it spins (rotational motion).
Step 2: The translational kinetic energy (TKE) is calculated using the formula TKE = (1/2)mv^2, where m is the mass of the object and v is its speed.
Step 3: The rotational kinetic energy (RKE) is calculated using the formula RKE = (1/2)Iω^2, where I is the moment of inertia and ω is the angular velocity.
Step 4: For a solid cylinder or sphere, the moment of inertia I can be expressed as (1/2)mR^2, and the angular velocity ω can be related to the speed v by the formula ω = v/R.
Step 5: Substitute ω into the RKE formula: RKE = (1/2)(1/2)mR^2(v/R)^2 = (1/4)mv^2.
Step 6: Now, add the translational kinetic energy and the rotational kinetic energy together: Total KE = TKE + RKE = (1/2)mv^2 + (1/4)mv^2.
Step 7: Combine the terms: (1/2)mv^2 + (1/4)mv^2 = (3/4)mv^2.
Step 8: Therefore, the total kinetic energy of the rolling object is (3/4)mv^2.
