Question

The L-shaped object in Figure 11-27 consists of three massesconnected by light rods.

Figure 11-27

(a) What torque must be applied to this object to giveit an angular acceleration of 1.21rad/s² if it is rotated about the x axis?
N·m
(b) What torque must be applied to this object to give it anangular acceleration of 1.21rad/s² if it is rotated about the y axis?
N·m
(c) What torque must be applied to this object to give it anangular acceleration of 1.21rad/s² if it is rotated about the z axis (whichis through the origin and perpendicular to the page)?
N·m

Answer 1

Concepts and reason

The concepts required to solve the given problem are moment of inertia and torque.

Initially, determine the moment of inertial each case by using expression for moment of inertia of a point mass. Then, determine the torque in each given case by using expression for torque in terms of moment of inertia and angular acceleration.

Fundamentals

The moment of inertia I of a point mass of mass m rotating about an axis and at distance r from the pivot is given by following expression.

$I = m{r^2}$

The torque $\tau$is defined as the product of moment of inertia I and angular acceleration $\alpha .$

$\tau = I\alpha$

(a)

The moment of inertia ${I_x}$of the system about x-axis is given by following expression.

${I_x} = {m_y}{y^2}$

Here, y is the distance to mass ${m_y}$on y-axis.

Substitute 9.0 kg for ${m_y}$, and 1.0 m for y in the above equation.

$\begin{array}{c}\\{I_x} = \left( {9.0\,{\rm{kg}}} \right){\left( {1.0\,{\rm{m}}} \right)^2}\\\\ = 9\,{\rm{kg}} \cdot {{\rm{m}}^2}\\\end{array}$

The torque ${\tau _x}$required to rotate about x-axis is,

${\tau _x} = {I_x}\alpha$

Substitute $9\,{\rm{kg}} \cdot {{\rm{m}}^2}$for ${I_x},$and $1.21\,{\rm{rad/}}{{\rm{s}}^2}$ for $\alpha$in the above equation.

$\begin{array}{c}\\{\tau _x} = \left( {9\,{\rm{kg}} \cdot {{\rm{m}}^2}} \right)\left( {1.21\,{\rm{rad/}}{{\rm{s}}^2}} \right)\\\\ = 10.89\,{\rm{N}} \cdot {\rm{m}}\\\end{array}$

(b)

The moment of inertia ${I_y}$of the system about y-axis is given by following expression.

${I_y} = {m_x}{x^2}$

Here, x is the distance to mass ${m_x}$on x-axis.

Substitute 2.5 kg for ${m_x}$, and 2.0 m for x in the above equation.

$\begin{array}{c}\\{I_y} = \left( {2.5\,{\rm{kg}}} \right){\left( {2.0} \right)^2}\\\\ = 10\,{\rm{kg}} \cdot {{\rm{m}}^2}\\\end{array}$

The torque ${\tau _y}$required to rotate about x-axis is,

${\tau _y} = {I_y}\alpha$

Substitute $10\,{\rm{kg}} \cdot {{\rm{m}}^2}$for ${I_x},$and $1.21\,{\rm{rad/}}{{\rm{s}}^2}$ for $\alpha$in the above equation.

$\begin{array}{c}\\{\tau _x} = \left( {10\,{\rm{kg}} \cdot {{\rm{m}}^2}} \right)\left( {1.21\,{\rm{rad/}}{{\rm{s}}^2}} \right)\\\\ = 12.1\,{\rm{N}} \cdot {\rm{m}}\\\end{array}$

(c)

The moment of inertia ${I_z}$of the system about z-axis is given by following expression.

${I_z} = {m_x}{x^2} + {m_y}{y^2}$

Substitute 2.5 kg for ${m_x}$, and 2.0 m for x, 9.0 kg for ${m_y},$and 1.0 m for y in the above equation.

$\begin{array}{c}\\{I_z} = \left( {2.5\,{\rm{kg}}} \right){\left( {2.0} \right)^2} + \left( {9.0\,{\rm{kg}}} \right){\left( {1.0\,{\rm{m}}} \right)^2}\\\\ = 10\,{\rm{kg}} \cdot {{\rm{m}}^2} + 9\,{\rm{kg}} \cdot {{\rm{m}}^2}\\\\ = 19\,{\rm{kg}} \cdot {{\rm{m}}^2}\\\end{array}$

The torque ${\tau _z}$required to rotate about z-axis is,

${\tau _z} = {I_z}\alpha$

Substitute $19\,{\rm{kg}} \cdot {{\rm{m}}^2}$for ${I_z},$and $1.21\,{\rm{rad/}}{{\rm{s}}^2}$ for $\alpha$in the above equation.

$\begin{array}{c}\\{\tau _x} = \left( {19\,{\rm{kg}} \cdot {{\rm{m}}^2}} \right)\left( {1.21\,{\rm{rad/}}{{\rm{s}}^2}} \right)\\\\ = 23\,{\rm{N}} \cdot {\rm{m}}\\\end{array}$

Ans: Part a

The torque required is $10.89\,{\rm{N}} \cdot {\rm{m}}{\rm{.}}$

Part b

The torque required is $12.1\,{\rm{N}} \cdot {\rm{m}}{\rm{.}}$

Part c

The torque required is $23\,{\rm{N}} \cdot {\rm{m}}{\rm{.}}$