Question

A uniform circular disc of radius 50cm at rest is free to turn about an axis which is perpendicular to its plane and passes through its centre. It is subjected to a torque which produces a constant angular acceleration of \[2.0\,{\text{rad}} \cdot {{\text{s}}^{ - 2}}\]. Its net acceleration in \[{\text{m}} \cdot {{\text{s}}^{ - 2}}\] at the end of \[2.0\,{\text{s}}\] is approximately:
A. 8.0
B. 7.0
C. 6.0
D. 3.0

Answer 1

Hint: Use the kinematic equation relating final angular velocity, initial angular velocity, angular acceleration and time to determine the final angular velocity of the disc at the end of \[2.0\,{\text{s}}\]. Determine the linear centripetal and tangential acceleration and then resultant of these two accelerations.

Formulae used:
The kinematic equation relating final angular velocity \[\omega \], initial angular velocity \[{\omega _0}\], angular acceleration \[\alpha \] and time \[t\] is
\[\omega = {\omega _0} + \alpha t\] …… (1)
The centripetal acceleration \[{a_C}\] of an object is
\[{a_C} = {\omega ^2}R\] …… (2)
Here, \[\omega \] is the angular velocity of the object and \[R\] is the radius of the circular path.
The tangential acceleration \[{a_t}\] of an object is
\[{a_t} = \alpha R\] …… (3)
Here, \[\alpha \] is the angular acceleration of the object and \[R\] is the radius of the circular path.

Complete step by step solution:
We have given that the radius of the disc is \[50\,{\text{cm}}\].
\[R = 50\,{\text{cm}}\]
The angular acceleration of the disc is \[2.0\,{\text{rad}} \cdot {{\text{s}}^{ - 2}}\].
\[\alpha = 2.0\,{\text{rad}} \cdot {{\text{s}}^{ - 2}}\]
The initial angular velocity of the object is zero as it starts from rest.
\[{\omega _0} = 0\,{\text{rads}} \cdot {{\text{s}}^{ - 1}}\]
Let us determine the angular velocity of the disc at the end of \[2.0\,{\text{s}}\] using equation (1).
Substitute \[0\,{\text{rads}} \cdot {{\text{s}}^{ - 1}}\] for \[{\omega _0}\], \[2.0\,{\text{rad}} \cdot {{\text{s}}^{ - 2}}\] for \[\alpha \] and \[2.0\,{\text{s}}\] for \[t\] in equation (1).
\[\omega = \left( {0\,{\text{rads}} \cdot {{\text{s}}^{ - 1}}} \right) + \left( {2.0\,{\text{rad}} \cdot {{\text{s}}^{ - 2}}} \right)\left( {2.0\,{\text{s}}} \right)\]
\[ \Rightarrow \omega = 4\,{\text{rad}} \cdot {{\text{s}}^{ - 1}}\]

Hence, the angular velocity of the disc at the end of \[2.0\,{\text{s}}\] is \[4\,{\text{rad}} \cdot {{\text{s}}^{ - 1}}\].
Let us determine the centripetal acceleration of the disc.
Substitute \[4\,{\text{rad}} \cdot {{\text{s}}^{ - 1}}\] for \[\omega \] and \[50\,{\text{cm}}\] for \[R\] in equation (2).
\[{a_C} = {\left( {4\,{\text{rad}} \cdot {{\text{s}}^{ - 1}}} \right)^2}\left( {50\,{\text{cm}}} \right)\]
\[ \Rightarrow {a_C} = {\left( {4\,{\text{rad}} \cdot {{\text{s}}^{ - 1}}} \right)^2}\left[ {\left( {50\,{\text{cm}}} \right)\left( {\dfrac{{{{10}^{ - 2}}\,{\text{m}}}}{{1\,{\text{cm}}}}} \right)} \right]\]
\[ \Rightarrow {a_C} = 8\,{\text{m/}}{{\text{s}}^2}\]
Let us determine the tangential acceleration of the disc.
Substitute \[2.0\,{\text{rad}} \cdot {{\text{s}}^{ - 2}}\] for \[\alpha \] and \[50\,{\text{cm}}\] for \[R\]in equation (3).
\[{a_t} = \left( {2.0\,{\text{rad}} \cdot {{\text{s}}^{ - 2}}} \right)\left( {50\,{\text{cm}}} \right)\]
\[ \Rightarrow {a_t} = \left( {2.0\,{\text{rad}} \cdot {{\text{s}}^{ - 2}}} \right)\left[ {\left( {50\,{\text{cm}}} \right)\left( {\dfrac{{{{10}^{ - 2}}\,{\text{m}}}}{{1\,{\text{cm}}}}} \right)} \right]\]
\[ \Rightarrow {a_t} = 1\,{\text{m/}}{{\text{s}}^2}\]
The net linear acceleration of the disc is the resultant of the centripetal and tangential acceleration of the disc.
\[a = \sqrt {a_C^2 + a_t^2} \]
Substitute \[8\,{\text{m/}}{{\text{s}}^2}\] for \[{a_C}\] and \[1\,{\text{m/}}{{\text{s}}^2}\] for \[{a_t}\] in the above equation.
\[a = \sqrt {{{\left( {8\,{\text{m/}}{{\text{s}}^2}} \right)}^2} + {{\left( {1\,{\text{m/}}{{\text{s}}^2}} \right)}^2}} \]
\[ \Rightarrow a = \sqrt {64 + 1} \]
\[ \Rightarrow a = \sqrt {65} \]
\[ \therefore a \approx 8\,{\text{m}} \cdot {{\text{s}}^{ - 2}}\]

Therefore, the net acceleration of the disc is approximately \[8\,{\text{m}} \cdot {{\text{s}}^{ - 2}}\].Hence, the correct option is A.

Note: The students may think that we have given the angular acceleration of the object and we need to determine the angular acceleration at the end of 2 seconds. But the student should read the question carefully because we have asked the net acceleration in meter per second square which is the unit of linear acceleration. So, we have to determine the linear acceleration at the end of 2 seconds.