# 8.5 Dynamics of rotational motion: rotational inertia  (Page 5/8)

 Page 5 / 8

Calculate the moment of inertia of a skater given the following information. (a) The 60.0-kg skater is approximated as a cylinder that has a 0.110-m radius. (b) The skater with arms extended is approximately a cylinder that is 52.5 kg, has a 0.110-m radius, and has two 0.900-m-long arms which are 3.75 kg each and extend straight out from the cylinder like rods rotated about their ends.

The triceps muscle in the back of the upper arm extends the forearm. This muscle in a professional boxer exerts a force of $\text{2.00}×{\text{10}}^{3}\phantom{\rule{0.25em}{0ex}}\text{N}$ with an effective perpendicular lever arm of 3.00 cm, producing an angular acceleration of the forearm of $\text{120}\phantom{\rule{0.25em}{0ex}}{\text{rad/s}}^{2}$ . What is the moment of inertia of the boxer’s forearm?

$\text{5.00 kg}\cdot {\text{m}}^{2}$

A soccer player extends her lower leg in a kicking motion by exerting a force with the muscle above the knee in the front of her leg. She produces an angular acceleration of $30.00 rad/{\text{s}}^{2}$ and her lower leg has a moment of inertia of $\text{0.750 kg}\cdot {\text{m}}^{2}$ . What is the force exerted by the muscle if its effective perpendicular lever arm is 1.90 cm?

Suppose you exert a force of 180 N tangential to a 0.280-m-radius 75.0-kg grindstone (a solid disk).

(a)What torque is exerted? (b) What is the angular acceleration assuming negligible opposing friction? (c) What is the angular acceleration if there is an opposing frictional force of 20.0 N exerted 1.50 cm from the axis?

(a) $50.4 N\cdot \text{m}$

(b) $\text{17.1}\phantom{\rule{0.25em}{0ex}}{\text{rad/s}}^{2}$

(c) $17.0\phantom{\rule{0.25em}{0ex}}{\text{rad/s}}^{2}$

Consider the 12.0 kg motorcycle wheel shown in [link] . Assume it to be approximately an annular ring with an inner radius of 0.280 m and an outer radius of 0.330 m. The motorcycle is on its center stand, so that the wheel can spin freely. (a) If the drive chain exerts a force of 2200 N at a radius of 5.00 cm, what is the angular acceleration of the wheel? (b) What is the tangential acceleration of a point on the outer edge of the tire? (c) How long, starting from rest, does it take to reach an angular velocity of 80.0 rad/s?

Zorch, an archenemy of Superman, decides to slow Earth’s rotation to once per 28.0 h by exerting an opposing force at and parallel to the equator. Superman is not immediately concerned, because he knows Zorch can only exert a force of $4.00×{\text{10}}^{7}\phantom{\rule{0.25em}{0ex}}\text{N}$ (a little greater than a Saturn V rocket’s thrust). How long must Zorch push with this force to accomplish his goal? (This period gives Superman time to devote to other villains.) Explicitly show how you follow the steps found in Problem-Solving Strategy for Rotational Dynamics .

$3\text{.}\text{96}×{\text{10}}^{\text{18}}\phantom{\rule{0.25em}{0ex}}\text{s}$

or $1.26×{\text{10}}^{\text{11}}\phantom{\rule{0.25em}{0ex}}\text{y}$

An automobile engine can produce 200 N ∙ m of torque. Calculate the angular acceleration produced if 95.0% of this torque is applied to the drive shaft, axle, and rear wheels of a car, given the following information. The car is suspended so that the wheels can turn freely. Each wheel acts like a 15.0 kg disk that has a 0.180 m radius. The walls of each tire act like a 2.00-kg annular ring that has inside radius of 0.180 m and outside radius of 0.320 m. The tread of each tire acts like a 10.0-kg hoop of radius 0.330 m. The 14.0-kg axle acts like a rod that has a 2.00-cm radius. The 30.0-kg drive shaft acts like a rod that has a 3.20-cm radius.

Starting with the formula for the moment of inertia of a rod rotated around an axis through one end perpendicular to its length , prove that the moment of inertia of a rod rotated about an axis through its center perpendicular to its length is . You will find the graphics in [link] useful in visualizing these rotations.

$\begin{array}{c}{I}_{\text{end}}={I}_{\text{center}}+m{\left(\frac{l}{2}\right)}^{2}\\ \text{Thus,}\phantom{\rule{0.25em}{0ex}}{I}_{\text{center}}={I}_{\text{end}}-\frac{1}{4}{\text{ml}}^{2}=\frac{1}{3}{\text{ml}}^{2}-\frac{1}{4}{\text{ml}}^{2}=\frac{1}{\text{12}}{\text{ml}}^{2}\end{array}$

Unreasonable Results

A gymnast doing a forward flip lands on the mat and exerts a 500-N ∙ m torque to slow and then reverse her angular velocity. Her initial angular velocity is 10.0 rad/s, and her moment of inertia is $0.050\phantom{\rule{0.25em}{0ex}}\text{kg}\cdot {\text{m}}^{2}$ . (a) What time is required for her to exactly reverse her spin? (b) What is unreasonable about the result? (c) Which premises are unreasonable or inconsistent?

(a) 2.00 ms

(b) The time interval is too short.

(c) The moment of inertia is much too small, by one to two orders of magnitude. A torque of $\text{500 N}\cdot \text{m}$ is reasonable.

Unreasonable Results

An advertisement claims that an 800-kg car is aided by its 20.0-kg flywheel, which can accelerate the car from rest to a speed of 30.0 m/s. The flywheel is a disk with a 0.150-m radius. (a) Calculate the angular velocity the flywheel must have if 95.0% of its rotational energy is used to get the car up to speed. (b) What is unreasonable about the result? (c) Which premise is unreasonable or which premises are inconsistent?

(a) 17,500 rpm

(b) This angular velocity is very high for a disk of this size and mass. The radial acceleration at the edge of the disk is>50,000 gs.

(c) Flywheel mass and radius should both be much greater, allowing for a lower spin rate (angular velocity).

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