36.2: Problems
- Page ID
- 92271
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Translational Problem
Consider the following translational problem: a body of mass \(m=3.0 \mathrm{~kg}\) is initally at rest; then a force of \(F=5.0 \mathrm{~N}\) is applied to it for time \(t=7.0\) seconds. What is the final velocity \(v\) of the body?
Solution. Given the force, we can find the acceleration; knowing the acceleration and time, we can find the velocity. The applicable equations are
\[F =m a \]
\[v =a t+v_{0} .\]
Solving Eq. \(\PageIndex{1}\) for \(a\) and substituting into Eq. \(\PageIndex{2}\), we have
\[v=\left(\frac{F}{m}\right) t+v_{0}\]
Substituting the given values of \(F, m\), and \(t\), and using \(v_{0}=0\), we have
\[v=\left(\frac{5.0 \mathrm{~N}}{3.0 \mathrm{~kg}}\right)(7.0 \mathrm{~s})\]
or
\[v=11.67 \mathrm{~m} / \mathrm{s}\]
Rotational Problem
Now consider the following similar rotational problem, which can be solved using the same method: a body of moment of inertia \(I=3.0 \mathrm{~kg} \mathrm{~m}^{2}\) is initially at rest (not rotating); then a torque of \(\tau=5.0 \mathrm{~N} \mathrm{~m}\) is applied to it for time \(t=7.0\) seconds. What is the final angular velocity \(\omega\) of the body?
Solution. Given the torque, we can find the angular acceleration; knowing the angular acceleration and time, we can find the angular velocity. The applicable equations are analogous to those used for the translational problem:
\[\tau =I \alpha \]
\[\omega =\alpha t+\omega_{0} .\]
Solving Eq. \(\PageIndex{6}\) for \(\alpha\) and substituting into Eq. \(\PageIndex{7}\), we have
\[\omega=\left(\frac{\tau}{I}\right) t+\omega_{0}\]
Substituting the given values of \(\tau, I\), and \(t\), and using \(\omega_{0}=0\), we have
\[\omega=\left(\frac{5.0 \mathrm{~N} \mathrm{~m}}{3.0 \mathrm{~kg} \mathrm{~m}}\right)(7.0 \mathrm{~s})\]
or
\[\omega=11.67 \mathrm{rad} / \mathrm{s}\]