
A child top is spun with an angular acceleration, \[\alpha =4{{t}^{3}}-3{{t}^{2}}+2t\]where t is in seconds and \[\alpha \]is in radians per second square. At t=0, the top has an angular velocity given by \[\omega =2\] rad/s and a reference line on it is at angular position \[{{\theta }_{0}}=1\] rad.
Statement 1; expression for angular velocity \[\omega =2+{{t}^{2}}-{{t}^{3}}+{{t}^{4}}\] rad/s
Statement 2; expression for angular position \[\theta =1+2t-3{{t}^{2}}+4{{t}^{3}}\]rad
(A) only statement 1 is true
(B) only statement 2 is true
(C) both of them are true
(D) None of them are true
Answer
568.8k+ views
Hint:We are given with angular acceleration and angular velocity at a given time. We need to find the value of angular velocity and angular position. We know angular position divided by time gives angular velocity which in turn divided by time gives angular acceleration. We can use calculus to find it easily.
Complete step by step answer:
\[\begin{align}
& \alpha =4{{t}^{3}}-3{{t}^{2}}+2t \\
& \dfrac{d\omega }{dt}=4{{t}^{3}}-3{{t}^{2}}+2t \\
& dw=(4{{t}^{3}}-3{{t}^{2}}+2t)dt \\
\end{align}\]
Now integrating both sides, since this is integration without the limits, we will have to use constant of integration.
\[\int{dw}=\int{(4{{t}^{3}}-3{{t}^{2}}+2t)dt}+C\]
\[w={{t}^{4}}-{{t}^{3}}+{{t}^{2}}+C\]
Now in order to find out the constant of integration we have to use initial conditions. Given at t=0, value of angular velocity is 2 rad/s. So, \[C=2\]. Putting it back we get,
\[w={{t}^{4}}-{{t}^{3}}+{{t}^{2}}+2\]
This is the same value as it is given in statement 1. Hence, statement 1 is correct.
Now we need to find an angular position.
\[\begin{align}
& \dfrac{d\theta }{dt}=\omega =2+{{t}^{2}}-{{t}^{3}}+{{t}^{4}} \\
& d\theta =(2+{{t}^{2}}-{{t}^{3}}+{{t}^{4}})dt \\
\end{align}\]
Integrating both sides we get
\[\theta =2t+\dfrac{{{t}^{3}}}{3}-\dfrac{{{t}^{4}}}{4}+\dfrac{{{t}^{5}}}{5}+C\]
Again, using initial conditions
C=1
Thus \[\theta =2t+\dfrac{{{t}^{3}}}{3}-\dfrac{{{t}^{4}}}{4}+\dfrac{{{t}^{5}}}{5}+1\]
This does not match statement 2. Thus, statement 2 is false.
Hence, the correct statement is (A)
Note:While doing integration if we have upper and lower limits then we do not have to use the constant of integration and this type of integration is called definite integration. But if there are no limits then we have to use a constant. The whole value is found out using initial conditions and such type of integration is called indefinite integration.
Complete step by step answer:
\[\begin{align}
& \alpha =4{{t}^{3}}-3{{t}^{2}}+2t \\
& \dfrac{d\omega }{dt}=4{{t}^{3}}-3{{t}^{2}}+2t \\
& dw=(4{{t}^{3}}-3{{t}^{2}}+2t)dt \\
\end{align}\]
Now integrating both sides, since this is integration without the limits, we will have to use constant of integration.
\[\int{dw}=\int{(4{{t}^{3}}-3{{t}^{2}}+2t)dt}+C\]
\[w={{t}^{4}}-{{t}^{3}}+{{t}^{2}}+C\]
Now in order to find out the constant of integration we have to use initial conditions. Given at t=0, value of angular velocity is 2 rad/s. So, \[C=2\]. Putting it back we get,
\[w={{t}^{4}}-{{t}^{3}}+{{t}^{2}}+2\]
This is the same value as it is given in statement 1. Hence, statement 1 is correct.
Now we need to find an angular position.
\[\begin{align}
& \dfrac{d\theta }{dt}=\omega =2+{{t}^{2}}-{{t}^{3}}+{{t}^{4}} \\
& d\theta =(2+{{t}^{2}}-{{t}^{3}}+{{t}^{4}})dt \\
\end{align}\]
Integrating both sides we get
\[\theta =2t+\dfrac{{{t}^{3}}}{3}-\dfrac{{{t}^{4}}}{4}+\dfrac{{{t}^{5}}}{5}+C\]
Again, using initial conditions
C=1
Thus \[\theta =2t+\dfrac{{{t}^{3}}}{3}-\dfrac{{{t}^{4}}}{4}+\dfrac{{{t}^{5}}}{5}+1\]
This does not match statement 2. Thus, statement 2 is false.
Hence, the correct statement is (A)
Note:While doing integration if we have upper and lower limits then we do not have to use the constant of integration and this type of integration is called definite integration. But if there are no limits then we have to use a constant. The whole value is found out using initial conditions and such type of integration is called indefinite integration.
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