149780
The angular speed of a motor wheel is increased from \(1200 \mathrm{rpm}\) to \(3120 \mathrm{rpm}\) in \(\mathbf{1 6}\) seconds. The angular acceleration of the motor wheel is
149781
The angular speed of a fly wheel moving with uniform angular acceleration changes from \(1200 \mathrm{rpm}\) to \(3120 \mathrm{rpm}\) in 16 seconds. The angular acceleration in \(\mathrm{rad} / \mathrm{s}^{2}\) is
149782
The angular velocity of a ceiling fan reduces to \(\mathbf{5 0 \%}\) after 36 rotations since it is switched off. Assuming uniform retardation, the number of rotations it further makes before coming to rest is
1 12
2 18
3 48
4 36
Explanation:
A From the third law of motion, the equation of circular motion is given below \(\omega^{2}=\omega_{\mathrm{o}}^{2}-2 \alpha \theta\) Angular velocity reduces \((50 \%), \omega=\frac{\omega_{0}}{2}\) So, \(\quad\left(\frac{\omega_{0}}{2}\right)^{2}=\omega_{0}^{2}-2 \alpha(2 \pi \mathrm{n}) \quad\left(\begin{array}{c}\because \theta=2 \pi \mathrm{n} \\ \mathrm{n}=36\end{array}\right)\) \(\alpha=\frac{3 \omega_{0}^{2}}{4 \times 4 \pi \times 36}\) Now let Fan completes total \(\mathrm{n}\) revolution From the starting to come to rest. \(0=\omega_{0}^{2}-2 \alpha(2 \pi \mathrm{n}) \Rightarrow \mathrm{n}=\left(\frac{\omega_{0}^{2}}{4 \alpha \pi}\right)\) Substituting the value of \(\alpha\) from equation (i), we get- \(\mathrm{n}=\frac{\omega_{0}^{2}}{4 \pi} \times \frac{4 \times 4 \pi \times 36}{3 \omega_{0}^{2}}=48 \text { revolution }\) Number of rotation \(=48-36 \Rightarrow 12\)
AP EAMCET-06.09.2021
Rotational Motion
149783
A body initially at rest, starts rotating with a uniform acceleration and covers \(100 \pi\) rad in the first 5 seconds. Its angular speed at the end of 5 seconds is
1 \(20 \pi \mathrm{rad.s}\)
2 \(30 \pi \mathrm{rad} . \mathrm{s}^{-1}\)
3 \(40 \pi \mathrm{rad} . \mathrm{s}^{-1}\)
4 \(10 \pi \mathrm{rad} . \mathrm{s}^{-1}\)
Explanation:
C Given, \(\omega_{\mathrm{i}}=0 \quad\) [start from rest] \(\theta=100 \pi \mathrm{rad}\) \(\mathrm{t}=5 \mathrm{sec}\) From using the second equation of angular of motion, \(\theta=\omega_{0} \mathrm{t}+\frac{1}{2} \alpha \mathrm{t}^{2}\) \(100 \pi=0+\frac{1}{2} \times \alpha \times 25\) \(\alpha=\frac{200 \pi}{25}\) \(\alpha=8 \pi \mathrm{rad} / \mathrm{sec}^{2}\) Now, to find angular speed, we use - \(\omega=\omega_{0}+\alpha \mathrm{t}\) \(\omega=0+8 \pi \times 5\) \(\omega=40 \pi \mathrm{rad} / \mathrm{sec}\)
149780
The angular speed of a motor wheel is increased from \(1200 \mathrm{rpm}\) to \(3120 \mathrm{rpm}\) in \(\mathbf{1 6}\) seconds. The angular acceleration of the motor wheel is
149781
The angular speed of a fly wheel moving with uniform angular acceleration changes from \(1200 \mathrm{rpm}\) to \(3120 \mathrm{rpm}\) in 16 seconds. The angular acceleration in \(\mathrm{rad} / \mathrm{s}^{2}\) is
149782
The angular velocity of a ceiling fan reduces to \(\mathbf{5 0 \%}\) after 36 rotations since it is switched off. Assuming uniform retardation, the number of rotations it further makes before coming to rest is
1 12
2 18
3 48
4 36
Explanation:
A From the third law of motion, the equation of circular motion is given below \(\omega^{2}=\omega_{\mathrm{o}}^{2}-2 \alpha \theta\) Angular velocity reduces \((50 \%), \omega=\frac{\omega_{0}}{2}\) So, \(\quad\left(\frac{\omega_{0}}{2}\right)^{2}=\omega_{0}^{2}-2 \alpha(2 \pi \mathrm{n}) \quad\left(\begin{array}{c}\because \theta=2 \pi \mathrm{n} \\ \mathrm{n}=36\end{array}\right)\) \(\alpha=\frac{3 \omega_{0}^{2}}{4 \times 4 \pi \times 36}\) Now let Fan completes total \(\mathrm{n}\) revolution From the starting to come to rest. \(0=\omega_{0}^{2}-2 \alpha(2 \pi \mathrm{n}) \Rightarrow \mathrm{n}=\left(\frac{\omega_{0}^{2}}{4 \alpha \pi}\right)\) Substituting the value of \(\alpha\) from equation (i), we get- \(\mathrm{n}=\frac{\omega_{0}^{2}}{4 \pi} \times \frac{4 \times 4 \pi \times 36}{3 \omega_{0}^{2}}=48 \text { revolution }\) Number of rotation \(=48-36 \Rightarrow 12\)
AP EAMCET-06.09.2021
Rotational Motion
149783
A body initially at rest, starts rotating with a uniform acceleration and covers \(100 \pi\) rad in the first 5 seconds. Its angular speed at the end of 5 seconds is
1 \(20 \pi \mathrm{rad.s}\)
2 \(30 \pi \mathrm{rad} . \mathrm{s}^{-1}\)
3 \(40 \pi \mathrm{rad} . \mathrm{s}^{-1}\)
4 \(10 \pi \mathrm{rad} . \mathrm{s}^{-1}\)
Explanation:
C Given, \(\omega_{\mathrm{i}}=0 \quad\) [start from rest] \(\theta=100 \pi \mathrm{rad}\) \(\mathrm{t}=5 \mathrm{sec}\) From using the second equation of angular of motion, \(\theta=\omega_{0} \mathrm{t}+\frac{1}{2} \alpha \mathrm{t}^{2}\) \(100 \pi=0+\frac{1}{2} \times \alpha \times 25\) \(\alpha=\frac{200 \pi}{25}\) \(\alpha=8 \pi \mathrm{rad} / \mathrm{sec}^{2}\) Now, to find angular speed, we use - \(\omega=\omega_{0}+\alpha \mathrm{t}\) \(\omega=0+8 \pi \times 5\) \(\omega=40 \pi \mathrm{rad} / \mathrm{sec}\)
149780
The angular speed of a motor wheel is increased from \(1200 \mathrm{rpm}\) to \(3120 \mathrm{rpm}\) in \(\mathbf{1 6}\) seconds. The angular acceleration of the motor wheel is
149781
The angular speed of a fly wheel moving with uniform angular acceleration changes from \(1200 \mathrm{rpm}\) to \(3120 \mathrm{rpm}\) in 16 seconds. The angular acceleration in \(\mathrm{rad} / \mathrm{s}^{2}\) is
149782
The angular velocity of a ceiling fan reduces to \(\mathbf{5 0 \%}\) after 36 rotations since it is switched off. Assuming uniform retardation, the number of rotations it further makes before coming to rest is
1 12
2 18
3 48
4 36
Explanation:
A From the third law of motion, the equation of circular motion is given below \(\omega^{2}=\omega_{\mathrm{o}}^{2}-2 \alpha \theta\) Angular velocity reduces \((50 \%), \omega=\frac{\omega_{0}}{2}\) So, \(\quad\left(\frac{\omega_{0}}{2}\right)^{2}=\omega_{0}^{2}-2 \alpha(2 \pi \mathrm{n}) \quad\left(\begin{array}{c}\because \theta=2 \pi \mathrm{n} \\ \mathrm{n}=36\end{array}\right)\) \(\alpha=\frac{3 \omega_{0}^{2}}{4 \times 4 \pi \times 36}\) Now let Fan completes total \(\mathrm{n}\) revolution From the starting to come to rest. \(0=\omega_{0}^{2}-2 \alpha(2 \pi \mathrm{n}) \Rightarrow \mathrm{n}=\left(\frac{\omega_{0}^{2}}{4 \alpha \pi}\right)\) Substituting the value of \(\alpha\) from equation (i), we get- \(\mathrm{n}=\frac{\omega_{0}^{2}}{4 \pi} \times \frac{4 \times 4 \pi \times 36}{3 \omega_{0}^{2}}=48 \text { revolution }\) Number of rotation \(=48-36 \Rightarrow 12\)
AP EAMCET-06.09.2021
Rotational Motion
149783
A body initially at rest, starts rotating with a uniform acceleration and covers \(100 \pi\) rad in the first 5 seconds. Its angular speed at the end of 5 seconds is
1 \(20 \pi \mathrm{rad.s}\)
2 \(30 \pi \mathrm{rad} . \mathrm{s}^{-1}\)
3 \(40 \pi \mathrm{rad} . \mathrm{s}^{-1}\)
4 \(10 \pi \mathrm{rad} . \mathrm{s}^{-1}\)
Explanation:
C Given, \(\omega_{\mathrm{i}}=0 \quad\) [start from rest] \(\theta=100 \pi \mathrm{rad}\) \(\mathrm{t}=5 \mathrm{sec}\) From using the second equation of angular of motion, \(\theta=\omega_{0} \mathrm{t}+\frac{1}{2} \alpha \mathrm{t}^{2}\) \(100 \pi=0+\frac{1}{2} \times \alpha \times 25\) \(\alpha=\frac{200 \pi}{25}\) \(\alpha=8 \pi \mathrm{rad} / \mathrm{sec}^{2}\) Now, to find angular speed, we use - \(\omega=\omega_{0}+\alpha \mathrm{t}\) \(\omega=0+8 \pi \times 5\) \(\omega=40 \pi \mathrm{rad} / \mathrm{sec}\)
149780
The angular speed of a motor wheel is increased from \(1200 \mathrm{rpm}\) to \(3120 \mathrm{rpm}\) in \(\mathbf{1 6}\) seconds. The angular acceleration of the motor wheel is
149781
The angular speed of a fly wheel moving with uniform angular acceleration changes from \(1200 \mathrm{rpm}\) to \(3120 \mathrm{rpm}\) in 16 seconds. The angular acceleration in \(\mathrm{rad} / \mathrm{s}^{2}\) is
149782
The angular velocity of a ceiling fan reduces to \(\mathbf{5 0 \%}\) after 36 rotations since it is switched off. Assuming uniform retardation, the number of rotations it further makes before coming to rest is
1 12
2 18
3 48
4 36
Explanation:
A From the third law of motion, the equation of circular motion is given below \(\omega^{2}=\omega_{\mathrm{o}}^{2}-2 \alpha \theta\) Angular velocity reduces \((50 \%), \omega=\frac{\omega_{0}}{2}\) So, \(\quad\left(\frac{\omega_{0}}{2}\right)^{2}=\omega_{0}^{2}-2 \alpha(2 \pi \mathrm{n}) \quad\left(\begin{array}{c}\because \theta=2 \pi \mathrm{n} \\ \mathrm{n}=36\end{array}\right)\) \(\alpha=\frac{3 \omega_{0}^{2}}{4 \times 4 \pi \times 36}\) Now let Fan completes total \(\mathrm{n}\) revolution From the starting to come to rest. \(0=\omega_{0}^{2}-2 \alpha(2 \pi \mathrm{n}) \Rightarrow \mathrm{n}=\left(\frac{\omega_{0}^{2}}{4 \alpha \pi}\right)\) Substituting the value of \(\alpha\) from equation (i), we get- \(\mathrm{n}=\frac{\omega_{0}^{2}}{4 \pi} \times \frac{4 \times 4 \pi \times 36}{3 \omega_{0}^{2}}=48 \text { revolution }\) Number of rotation \(=48-36 \Rightarrow 12\)
AP EAMCET-06.09.2021
Rotational Motion
149783
A body initially at rest, starts rotating with a uniform acceleration and covers \(100 \pi\) rad in the first 5 seconds. Its angular speed at the end of 5 seconds is
1 \(20 \pi \mathrm{rad.s}\)
2 \(30 \pi \mathrm{rad} . \mathrm{s}^{-1}\)
3 \(40 \pi \mathrm{rad} . \mathrm{s}^{-1}\)
4 \(10 \pi \mathrm{rad} . \mathrm{s}^{-1}\)
Explanation:
C Given, \(\omega_{\mathrm{i}}=0 \quad\) [start from rest] \(\theta=100 \pi \mathrm{rad}\) \(\mathrm{t}=5 \mathrm{sec}\) From using the second equation of angular of motion, \(\theta=\omega_{0} \mathrm{t}+\frac{1}{2} \alpha \mathrm{t}^{2}\) \(100 \pi=0+\frac{1}{2} \times \alpha \times 25\) \(\alpha=\frac{200 \pi}{25}\) \(\alpha=8 \pi \mathrm{rad} / \mathrm{sec}^{2}\) Now, to find angular speed, we use - \(\omega=\omega_{0}+\alpha \mathrm{t}\) \(\omega=0+8 \pi \times 5\) \(\omega=40 \pi \mathrm{rad} / \mathrm{sec}\)
