a lawn mower has a flat, rod-shaped steel blade that rotates about its center. the mass of the blade is $0.65\text{ kg}$ and its length is $0.55\text{ m}$.

part a
what is the rotational energy of the blade at its operating angular speed of $3530\text{ rpm}$?
express your answer in joules.
$k_r = $

part b
if all of the rotational kinetic energy of the blade could be converted to gravitational potential energy, to what height would the blade rise?
express your answer in meters.
$y = $

Question

a lawn mower has a flat, rod-shaped steel blade that rotates about its center. the mass of the blade is $0.65\text{ kg}$ and its length is $0.55\text{ m}$.

part a
what is the rotational energy of the blade at its operating angular speed of $3530\text{ rpm}$?
express your answer in joules.
$k_r = $

part b
if all of the rotational kinetic energy of the blade could be converted to gravitational potential energy, to what height would the blade rise?
express your answer in meters.
$y = $

Accepted Answer

{ "quality": "clear", "question_count": 2, "skills_matched": [ "step_cot", "explain_then_answer", "quick" ], "needs_drawing": false, "concepts_used": [ "Moment of Inertia", "Rotational Kinematics", "Conservation of Energy" ], "new_concepts": [ "Rotational Kinetic Energy" ], "current_concepts": [ "Moment of Inertia", "Rotational Kinematics", "Rotational Kinetic Energy", "Conservation of Energy" ] } ## Calculate the moment of inertia of the rod-shaped blade $$ \begin{aligned} m &= 0.65\text{ kg}\ L &= 0.55\text{ m}\ I &= \frac{1}{12} m L^2 = \frac{1}{12} (0.65\text{ kg}) (0.55\text{ m})^2 \approx 0.0163906\text{ kg}\cdot\text{m}^2 \end{aligned} $$ ## Convert angular speed to radians per second and calculate rotational energy $$ \begin{aligned} \omega &= 3530\text{ rpm} = 3530 \times \frac{2\pi}{60}\text{ rad/s} \approx 369.66\text{ rad/s}\ K_r &= \frac{1}{2} I \omega^2 = \frac{1}{2} (0.0163906\text{ kg}\cdot\text{m}^2) (369.66\text{ rad/s})^2 \approx 1120\text{ J} \end{aligned} $$ ## Calculate the height of rise using conservation of energy $$ \begin{aligned} g &= 9.80\text{ m/s}^2\ K_r &= m g y\ y &= \frac{K_r}{m g} = \frac{1120\text{ J}}{(0.65\text{ kg})(9.80\text{ m/s}^2)} \approx 176\text{ m} \end{aligned} $$ ## Question 1 $K_r =$ 1120 J ## Question 2 $y =$ 176 m { "subject": "Natural Science", "question_type": "Multi-part", "knowledge_point": [ "Natural Science", "Physics", "Rotational Kinetic Energy" ] }

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QUESTION IMAGE

Question

Explanation:

Calculate the moment of inertia of the rod-shaped blade

Convert angular speed to radians per second and calculate rotational energy

Calculate the height of rise using conservation of energy

Answer:

Question 1

Question 2