Respuesta :
Deceleration due to rolling resistance (α) = coefficient*g/R = 0.1*9.81/0.5 = 1.962 rad/sec^2
Equations of motion:
ωf=ωo+αt
But ωf = 0 rad/sec; ωo= 2*pi*RPM/60 = 4.19 rad/sec
Therefore,
0=4.19-1.962t => 1.962t = 4.19 => t = 4.19/1.962 = 2.13 seconds
Equations of motion:
ωf=ωo+αt
But ωf = 0 rad/sec; ωo= 2*pi*RPM/60 = 4.19 rad/sec
Therefore,
0=4.19-1.962t => 1.962t = 4.19 => t = 4.19/1.962 = 2.13 seconds
The train with a constant angular deceleration will stop in [tex]\fbox{2.13\text{ s}}[/tex].
Further Explanation:
An object in pure rolling motion on a rough surface experiences resistive rolling force.
Given:
The diameter of wheel is [tex]1\,{\text{m}}[/tex].
The angular speed is [tex]40\,{\text{rpm}}[/tex].
Concept:
The equation of resistive or friction force is:
[tex]\fbox{\begin\\F_r=m\cdot\text{a}_r\end{minispace}}[/tex]
Here, [tex]m[/tex] is the mass of body, [tex]{a_r}[/tex] is the deceleration due to rolling resistance and [tex]{F_r}[/tex] is the frictional force.
The equation of normal force is:
[tex]\fbox{N = mg}[/tex]
Here, [tex]g[/tex] is the gravitational acceleration, [tex]m[/tex] is the mass of body, and [tex]N[/tex] is the normal force (which is equal to weight of body).
For body to be in equilibrium the equation of forces is:
[tex]{F_r} = \dfrac{{bN}}{R}[/tex]
Here, [tex]b[/tex] is the coefficient of rolling friction and [tex]R[/tex] is the radius of wheel.
Substitute [tex]m{a_r}[/tex] for [tex]{F_r}[/tex] and [tex]mg[/tex] for [tex]N[/tex] in above equation.
[tex]\begin{gathered}m{a_r}=\frac{{bmg}}{R}\\{a_r}=\frac{{bg}}{R}\\\end{gathered}[/tex]
Here, [tex]{a_r}[/tex] is the deceleration due to rolling resistance, [tex]b[/tex] is the coefficient of rolling friction, [tex]g[/tex] is the acceleration due to gravity and [tex]R[/tex] is the radius of wheel.
Substitute [tex]0.1[/tex] for [tex]b[/tex], [tex]9.81\text{ m}/\text{s}^2[/tex] for [tex]g[/tex], and [tex]\dfrac{1}{2}[/tex] for [tex]R[/tex] in above equation.
[tex]\begin{gathered}{a_r}=\frac{{0.1\times9.81\,{{\text{m}}\mathord{\left/{\vphantom{{\text{m}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}}}{{0.5}}\\=1.962\,{{{\text{rad}}}\mathord{\left/{\vphantom{{{\text{rad}}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}\\\end{gathered}[/tex]
The initial angular velocity of body is:
[tex]{\omega _0} = \dfrac{{2\pi N}}{{60}}[/tex]
Here, [tex]N[/tex] is the angular speed in rpm and [tex]{\omega _0}[/tex] is the initial angular velocity.
Substitute [tex]40\,{\text{rpm}}[/tex] for [tex]N[/tex] in above equation.
[tex]\begin{gathered}{\omega _0}=\frac{{2\pi\times40\,{\text{rpm}}}}{{60}}\\=4.19\,{{{\text{rad}}}\mathord{\left/{\vphantom{{{\text{rad}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}\\\end{gathered}[/tex]
Applying equation of motion:
[tex]{\omega _f}={\omega _0}-{\alpha _r}t[/tex]
Rearrange the above equation for value t:
[tex]t=\dfrac{{{\omega _0}-{\omega _f}}}{{{\alpha _r}}}[/tex]
Here, [tex]{\omega _f}[/tex] is the final velocity, [tex]{\omega _0}[/tex] is the initial velocity, [tex]{\alpha _r}[/tex] is the deceleration and [tex]t[/tex] is the time taken by train to come to rest.
Substitute [tex]0\,{{{\text{rad}}}\mathord{\left/{\vphantom{{{\text{rad}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}[/tex] for [tex]{\omega _f}[/tex] and [tex]1.962\,{{{\text{rad}}} \mathord{\left/ {\vphantom {{{\text{rad}}} {{{\text{s}}^{\text{2}}}}}} \right. \kern-\nulldelimiterspace} {{{\text{s}}^{\text{2}}}}}[/tex] for [tex]{\alpha _r}[/tex] and [tex]4.19\,{{{\text{rad}}} \mathord{\left/ {\vphantom {{{\text{rad}}} {\text{s}}}} \right. \kern-\nulldelimiterspace} {\text{s}}}[/tex] for [tex]{\omega _0}[/tex] in above equation.
[tex]\begin{aligned}\\t&=\frac{{4.19\,{{{\text{rad}}}\mathord{\left/{\vphantom{{{\text{rad}}}{\text{s}}}}\right.\kern-\nulldelimiterspace}{\text{s}}}-0}}{{1.962\,{{{\text{rad}}}\mathord{\left/{\vphantom{{{\text{rad}}}{{{\text{s}}^{\text{2}}}}}}\right.\kern-\nulldelimiterspace}{{{\text{s}}^{\text{2}}}}}}}\\&=2.13\,{\text{s}}\\\end{aligned}[/tex]
Thus, the train with a constant angular deceleration will stop in [tex]\fbox{2.13\,{\text{s}}}[/tex].
Learn more:
1. Motion of a block under friction https://brainly.com/question/7031524
2. Conservation of momentum in collision https://brainly.com/question/9484203
3. Motion of a ball under gravity https://brainly.com/question/10934170
Answer Details:
Grade: High School
Subject: Physics
Chapter: Kinematics
Keywords:
Horizontal track, angular speed, coefficient of rolling friction, 40 rpm, and deceleration.
