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Chapter 10: Problem 40

A hollow, thin-walled sphere of mass 12.0 \(\mathrm{kg}\) and diameter 48.0 \(\mathrm{cm}\) is rotating about an axle through its center. The angle (in radians) through which it turns as a function of time (in seconds) is given by \(\theta(t)=A t^{2}+B t^{4},\) where \(A\) has numerical value 1.50 and \(B\) has numerical value 1.10 . (a) What are the units of the constants \(A\) and \(B ?\) (b) At the time 3.00 s, find (i) the angular momentum of the sphere and (ii) the net torque on the sphere.

Short Answer

Expert verified
(a) A: \(1/s^2\), B: \(1/s^4\); (b) (i) 58.88 kg路m虏/s, (ii) 56.10 N路m

Step by step solution

01

Determine Units for A and B

The function for the angle is \(\theta(t) = A t^2 + B t^4\). Since \(\theta(t)\) is in radians, a dimensionless unit, the units for \(A\cdot t^2\) and \(B\cdot t^4\) must also result in radians (dimensionless). This gives:- For \(A\), units must be \(\text{radians}/t^2\), which is \(1/s^2\).- For \(B\), units must be \(\text{radians}/t^4\), which is \(1/s^4\).
02

Calculate Angular Velocity

The angular velocity \(\omega(t)\) is the first derivative of \(\theta(t)\) with respect to time \(t\). \[ \omega(t) = \frac{d\theta}{dt} = \frac{d}{dt}(A t^2 + B t^4) = 2At + 4Bt^3 \]Substitute \(A = 1.50\) and \(B = 1.10\) to find \(\omega(3.00\,\text{s})\): \[ \omega(3.00) = 2 \times 1.50 \times 3.00 + 4 \times 1.10 \times (3.00)^3 = 9.00 + 118.80 = 127.80\, \text{rad/s} \]
03

Determine Moment of Inertia

For a hollow, thin-walled sphere, the moment of inertia \(I\) is given by \(I = \frac{2}{3}m r^2\), where \(m = 12.0\,\text{kg}\) and the radius \(r = \frac{48.0}{2} = 24.0\,\text{cm} = 0.24\,\text{m}\). \[ I = \frac{2}{3} \times 12.0 \times (0.24)^2 = 0.4608\,\text{kg}\cdot \text{m}^2 \]
04

Calculate Angular Momentum

The angular momentum \(L\) is given by \(L = I \omega\). Substitute \(I = 0.4608\, \text{kg}\cdot \text{m}^2\) and \(\omega(3.00) = 127.80\, \text{rad/s}\): \[ L = 0.4608 \times 127.80 = 58.88\, \text{kg}\cdot \text{m}^2/\text{s} \]
05

Determine Net Torque

Net torque \(\tau\) is the time rate of change of angular momentum. We find \(\tau\) using the relation \(\tau = I \frac{d\omega}{dt}\), where \(\frac{d\omega}{dt} = \frac{d}{dt}(2At + 4Bt^3) = 2A + 12Bt^2\).For \(t = 3.00\,\text{s}\), substituting \(A = 1.50\) and \(B = 1.10\): \[ \frac{d\omega}{dt}(3.00) = 2 \times 1.50 + 12 \times 1.10 \times 3.00^2 = 3.00 + 118.80 = 121.80\, \text{rad/s}^2 \]The net torque \(\tau\) is then: \[ \tau = 0.4608 \times 121.80 = 56.10\, \text{N}\cdot \text{m} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Moment of Inertia
The moment of inertia is a crucial concept when discussing rotational dynamics. It represents how mass is distributed in relation to an axis of rotation.
To picture it, consider how difficult it is to spin a heavy dumbbell around its midpoint compared to spinning it around one of its ends. This resistance to rotational motion is what we call the moment of inertia.For a hollow, thin-walled sphere, the moment of inertia is calculated using the formula: - \[ I = \frac{2}{3} m r^2 \]- where \( m \) is the mass of the sphere and \( r \) is the radius.Its unit is the kilogram meter squared (\( \text{kg} \cdot \text{m}^2 \)).
In our example, with a sphere of **mass 12 kg** and **radius 0.24 m**, the moment of inertia calculates to approximately **0.4608 kg路m虏**. This value indicates how much the sphere resists angular acceleration when a torque is applied.
Angular Velocity
Angular velocity describes how fast something is rotating. Instead of measuring speed in meters per second (as we do in linear motion), angular velocity is measured in radians per second.
To find angular velocity from an angular displacement function, you take the derivative concerning time. If you think of angular velocity as the rotational counterpart to linear velocity, it helps visualize how quickly a rotating object changes its position.In our context:- The given angular displacement function is \( \theta(t) = A t^2 + B t^4 \).- To find angular velocity \( \omega(t) \), we differentiate this to obtain \( \omega(t) = 2At + 4Bt^3 \).Plugging in the numbers \( A = 1.50 \) and \( B = 1.10 \) at \( t = 3 \) seconds, the angular velocity comes out as **127.80 rad/s**. This value shows how fast the sphere is turning at that moment.
Net Torque
Net torque is essential for understanding the change in rotational motion. It's the rotational equivalent of force in linear motion and is quantified in Newton-meters (\( \text{N} \cdot \text{m} \)).
Torque is what causes objects to rotate and determine the rate of change of angular momentum.The formula for net torque \( \tau \) is derived from:- \[ \tau = I \frac{d\omega}{dt} \]- where \( \frac{d\omega}{dt} \) is the angular acceleration.In practice, angular acceleration can be found by differentiating angular velocity \( \omega(t) = 2At + 4Bt^3 \). Doing so gives us:- \( \frac{d\omega}{dt} = 2A + 12Bt^2 \).Substitute \( A = 1.50 \) and \( B = 1.10 \) at \( t = 3 \, \text{s} \). We achieve an angular acceleration of **121.80 rad/s虏**. When combined with the moment of inertia \( I = 0.4608 \, \text{kg} \cdot \text{m}^2 \), we find the net torque **56.10 N路m**. This result indicates the influence required to alter the sphere's rotational speed over time.

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Most popular questions from this chapter

Tarzan and Jane in the 21 st Century. Tarzan has foolishly gotten himself into another scrape with the animals and must be rescued once again by Jane. The 60.0 -kg Jane starts from rest at a height of 5.00 \(\mathrm{m}\) in the trees and swings down to the ground using a thin, but very rigid, 30.0 -kg vine 8.00 \(\mathrm{m}\) long. She arrives just in time to snatch the 72.0 -kg Tarzan from the jaws of an angry hippopotamus. What is Jane's (and the vine's) angular speed (a) just before she grabs Tarzan and (b) just after she grabs him? (c) How high will Tarzan and Jane go on their first swing after this daring rescue?

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