A large roller in the form of a uniform cylinder is pulled...

A large roller in the form of a uniform cylinder is pulled by a tractor to compact earth; it has a 1.80-m diameter and weighs $10 \mathrm{kN}$. If frictional losses can be ignored, what average horsepower must the tractor provide to accelerate the cylinder from rest to a speed of $4.0 \mathrm{~m} / \mathrm{s}$ in a horizontal distance of $3.0 \mathrm{~m}$ ? The power required is equal to the work done by the tractor divided by the time it takes. The tractor does the following work: $$ \text { Work }=(\Delta \mathrm{KE})_{r}+(\Delta \mathrm{KE})_{t}=\frac{1}{2} I \omega_{f}^{2}+\frac{1}{2} m v_{f}^{2} $$ We have $v_{f}=4.0 \mathrm{~m} / \mathrm{s}, \omega_{f}=v_{f} / r=4.44 \mathrm{rad} / \mathrm{s}$, and $m=10000 / 9.81=1019 \mathrm{~kg} .$ The moment of inertia of the cylinder is $$ I=\frac{1}{2} m r^{2}=\frac{1}{2}(1019 \mathrm{~kg})(0.90 \mathrm{~m})^{2}=413 \mathrm{~kg} \cdot \mathrm{m}^{2} $$ Substituting these values, the work required tums out to be $12.23 \mathrm{~kJ}$. We still need the time taken to do this work. Because the roller went $3.0 \mathrm{~m}$ with an average velocity $v_{\mathrm{av}}=\frac{1}{2}(4+0)=2.0 \mathrm{~m} / \mathrm{s}$ $$ t=\frac{s}{v_{a v}}=\frac{3.0 \mathrm{~m}}{2.0 \mathrm{~m} / \mathrm{s}}=1.5 \mathrm{~s} $$ Then $\quad$ Power $=\frac{\text { Work }}{\text { Time }}=\frac{12230 \mathrm{~J}}{1.5 \mathrm{~s}}=(8150 \mathrm{~W})\left(\frac{1 \mathrm{hp}}{746 \mathrm{~W}}\right)=11 \mathrm{hp}$

A large roller in the form of a uniform cylinder is pulled by a tractor to compact earth; it has a 1.80-m diameter and weighs $10 \mathrm{kN}$. If frictional losses can be ignored, what average horsepower must the tractor provide to accelerate the cylinder from rest to a speed of $4.0 \mathrm{~m} / \mathrm{s}$ in a horizontal distance of $3.0 \mathrm{~m}$ ?

The power required is equal to the work done by the tractor divided by the time it takes. The tractor does the following work:
$$
\text { Work }=(\Delta \mathrm{KE})_{r}+(\Delta \mathrm{KE})_{t}=\frac{1}{2} I \omega_{f}^{2}+\frac{1}{2} m v_{f}^{2}
$$
We have $v_{f}=4.0 \mathrm{~m} / \mathrm{s}, \omega_{f}=v_{f} / r=4.44 \mathrm{rad} / \mathrm{s}$, and $m=10000 / 9.81=1019 \mathrm{~kg} .$ The moment of inertia of
the cylinder is
$$
I=\frac{1}{2} m r^{2}=\frac{1}{2}(1019 \mathrm{~kg})(0.90 \mathrm{~m})^{2}=413 \mathrm{~kg} \cdot \mathrm{m}^{2}
$$
Substituting these values, the work required tums out to be $12.23 \mathrm{~kJ}$.
We still need the time taken to do this work. Because the roller went $3.0 \mathrm{~m}$ with an average velocity $v_{\mathrm{av}}=\frac{1}{2}(4+0)=2.0 \mathrm{~m} / \mathrm{s}$
$$
t=\frac{s}{v_{a v}}=\frac{3.0 \mathrm{~m}}{2.0 \mathrm{~m} / \mathrm{s}}=1.5 \mathrm{~s}
$$
Then $\quad$ Power $=\frac{\text { Work }}{\text { Time }}=\frac{12230 \mathrm{~J}}{1.5 \mathrm{~s}}=(8150 \mathrm{~W})\left(\frac{1 \mathrm{hp}}{746 \mathrm{~W}}\right)=11 \mathrm{hp}$

Key Concepts

Translational Kinetic Energy

Translational kinetic energy is the energy an object possesses due to its motion along a path. It is given by one half the mass of the object multiplied by the square of its velocity, reflecting the energy required to accelerate the object linearly from rest.

Power

Power describes the rate at which work is done or energy is transferred over time. It provides a measure of how quickly energy changes occur in a system and is expressed in units such as watts or horsepower, making it essential for evaluating the performance of mechanical systems.

Kinematics

Kinematics is the branch of mechanics concerned with describing the motion of objects through displacement, velocity, acceleration, and time without accounting for the forces causing the motion. It is used to determine factors like the time required for an object to cover a certain distance under acceleration.

Angular Velocity

Angular velocity measures the rate of change of an object's rotational angle per unit time. It is fundamental in relating linear speeds to rotational motion and is directly used to compute rotational kinetic energy in a system.

Work

Work is the process of energy transfer when a force is applied over a distance. In mechanics, it is calculated as the product of the force component along the displacement and the displacement itself, and it quantifies how much energy is put into or taken from a system.

Rotational Kinetic Energy

Rotational kinetic energy is the energy an object has due to its rotation about an axis. It is expressed as one half the moment of inertia times the square of the angular velocity, and it represents the energy needed to spin the object from rest to its rotational speed.

Moment of Inertia

The moment of inertia quantifies an object's resistance to angular acceleration, analogous to mass in linear motion. It depends on the object's mass distribution relative to the axis of rotation, and it plays a crucial role in determining the rotational kinetic energy.

Transcript

Please give Ace some feedback