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Rotational Motion

149780 The angular speed of a motor wheel is increased from $1200 rpm$ to $3120 rpm$ in $16$ seconds. The angular acceleration of the motor wheel is

1

6 π rad / s^{2}

2

8 π rad / s^{2}

3

2 π rad / s^{2}

4

4 π rad / s^{2}

Explanation:

D Given,
$N_{1} = 1200 rpm$
$N_{2} = 3120 rpm$
$t = 16 \sec$
Initial angular speed-
$ω_{i} = \frac{2 π N_{1}}{60} = \frac{2 π \times 1200}{60} = 40 π rad / s$
Final angular speed-
$ω_{f} = \frac{2 π N_{2}}{60} = \frac{2 π \times 3120}{60} = 104 π rad / s$
So, Angular acceleration-
$α = \frac{(ω_{f} - ω_{i})}{t}$
$= \frac{(104 - 40) π}{16}$
$= 4 π rad / s^{2}$

Karnataka CET-2022

Rotational Motion

149781 The angular speed of a fly wheel moving with uniform angular acceleration changes from $1200 rpm$ to $3120 rpm$ in 16 seconds. The angular acceleration in $rad / s^{2}$ is

1

12 π

2

104 π

3

2 π

4

4 π

Explanation:

D Given,
Angular acceleration changes from $1200 rpm$ to 3120 rpm in $16 \sec$ .
Angular acceleration
$α = \frac{ω_{f} - ω_{i}}{t} From equation ω_{f} = ω_{i} + α t]$
$ω_{f} = 3120 \times \frac{2 π}{60} rad / \sec$
$ω_{f} = 104 π rad / \sec$
$ω_{i} = 1200 \times \frac{2 π}{60} rad / \sec$
$ω_{i} = 40 π rad / \sec$
Then,
$α = \frac{104 π - 40 π}{16} = \frac{64 π}{16}$
$α = 4 π rad / \sec^{2}$

NEET-17.07.2022

Rotational Motion

149782 The angular velocity of a ceiling fan reduces to $50 %$ 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
$ω^{2} = ω_{o}^{2} - 2 α θ$
Angular velocity reduces $(50 %), ω = \frac{ω_{0}}{2}$
So, ${(\frac{ω_{0}}{2})}^{2} = ω_{0}^{2} - 2 α (2 π n) (\begin{matrix} ∵ θ = 2 π n \\ n = 36 \end{matrix})$
$α = \frac{3 ω_{0}^{2}}{4 \times 4 π \times 36}$
Now let Fan completes total $n$ revolution
From the starting to come to rest.
$0 = ω_{0}^{2} - 2 α (2 π n) \Rightarrow n = (\frac{ω_{0}^{2}}{4 α π})$
Substituting the value of $α$ from equation (i), we get-
$n = \frac{ω_{0}^{2}}{4 π} \times \frac{4 \times 4 π \times 36}{3 ω_{0}^{2}} = 48 revolution$
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 $100 π$ rad in the first 5 seconds. Its angular speed at the end of 5 seconds is

1

20 π rad . s

2

30 π rad . s^{- 1}

3

40 π rad . s^{- 1}

4

10 π rad . s^{- 1}

Explanation:

C Given, $ω_{i} = 0$ [start from rest]
$θ = 100 π rad$
$t = 5 \sec$
From using the second equation of angular of motion,
$θ = ω_{0} t + \frac{1}{2} α t^{2}$
$100 π = 0 + \frac{1}{2} \times α \times 25$
$α = \frac{200 π}{25}$
$α = 8 π rad / \sec^{2}$
Now, to find angular speed, we use -
$ω = ω_{0} + α t$
$ω = 0 + 8 π \times 5$
$ω = 40 π rad / \sec$

AP EAMCET-03.09.2021

Rotational Motion

149780 The angular speed of a motor wheel is increased from $1200 rpm$ to $3120 rpm$ in $16$ seconds. The angular acceleration of the motor wheel is

1

6 π rad / s^{2}

2

8 π rad / s^{2}

3

2 π rad / s^{2}

4

4 π rad / s^{2}

Explanation:

D Given,
$N_{1} = 1200 rpm$
$N_{2} = 3120 rpm$
$t = 16 \sec$
Initial angular speed-
$ω_{i} = \frac{2 π N_{1}}{60} = \frac{2 π \times 1200}{60} = 40 π rad / s$
Final angular speed-
$ω_{f} = \frac{2 π N_{2}}{60} = \frac{2 π \times 3120}{60} = 104 π rad / s$
So, Angular acceleration-
$α = \frac{(ω_{f} - ω_{i})}{t}$
$= \frac{(104 - 40) π}{16}$
$= 4 π rad / s^{2}$

Karnataka CET-2022

Rotational Motion

149781 The angular speed of a fly wheel moving with uniform angular acceleration changes from $1200 rpm$ to $3120 rpm$ in 16 seconds. The angular acceleration in $rad / s^{2}$ is

1

12 π

2

104 π

3

2 π

4

4 π

Explanation:

D Given,
Angular acceleration changes from $1200 rpm$ to 3120 rpm in $16 \sec$ .
Angular acceleration
$α = \frac{ω_{f} - ω_{i}}{t} From equation ω_{f} = ω_{i} + α t]$
$ω_{f} = 3120 \times \frac{2 π}{60} rad / \sec$
$ω_{f} = 104 π rad / \sec$
$ω_{i} = 1200 \times \frac{2 π}{60} rad / \sec$
$ω_{i} = 40 π rad / \sec$
Then,
$α = \frac{104 π - 40 π}{16} = \frac{64 π}{16}$
$α = 4 π rad / \sec^{2}$

NEET-17.07.2022

Rotational Motion

149782 The angular velocity of a ceiling fan reduces to $50 %$ 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
$ω^{2} = ω_{o}^{2} - 2 α θ$
Angular velocity reduces $(50 %), ω = \frac{ω_{0}}{2}$
So, ${(\frac{ω_{0}}{2})}^{2} = ω_{0}^{2} - 2 α (2 π n) (\begin{matrix} ∵ θ = 2 π n \\ n = 36 \end{matrix})$
$α = \frac{3 ω_{0}^{2}}{4 \times 4 π \times 36}$
Now let Fan completes total $n$ revolution
From the starting to come to rest.
$0 = ω_{0}^{2} - 2 α (2 π n) \Rightarrow n = (\frac{ω_{0}^{2}}{4 α π})$
Substituting the value of $α$ from equation (i), we get-
$n = \frac{ω_{0}^{2}}{4 π} \times \frac{4 \times 4 π \times 36}{3 ω_{0}^{2}} = 48 revolution$
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 $100 π$ rad in the first 5 seconds. Its angular speed at the end of 5 seconds is

1

20 π rad . s

2

30 π rad . s^{- 1}

3

40 π rad . s^{- 1}

4

10 π rad . s^{- 1}

Explanation:

C Given, $ω_{i} = 0$ [start from rest]
$θ = 100 π rad$
$t = 5 \sec$
From using the second equation of angular of motion,
$θ = ω_{0} t + \frac{1}{2} α t^{2}$
$100 π = 0 + \frac{1}{2} \times α \times 25$
$α = \frac{200 π}{25}$
$α = 8 π rad / \sec^{2}$
Now, to find angular speed, we use -
$ω = ω_{0} + α t$
$ω = 0 + 8 π \times 5$
$ω = 40 π rad / \sec$

AP EAMCET-03.09.2021

Rotational Motion

149780 The angular speed of a motor wheel is increased from $1200 rpm$ to $3120 rpm$ in $16$ seconds. The angular acceleration of the motor wheel is

1

6 π rad / s^{2}

2

8 π rad / s^{2}

3

2 π rad / s^{2}

4

4 π rad / s^{2}

Explanation:

D Given,
$N_{1} = 1200 rpm$
$N_{2} = 3120 rpm$
$t = 16 \sec$
Initial angular speed-
$ω_{i} = \frac{2 π N_{1}}{60} = \frac{2 π \times 1200}{60} = 40 π rad / s$
Final angular speed-
$ω_{f} = \frac{2 π N_{2}}{60} = \frac{2 π \times 3120}{60} = 104 π rad / s$
So, Angular acceleration-
$α = \frac{(ω_{f} - ω_{i})}{t}$
$= \frac{(104 - 40) π}{16}$
$= 4 π rad / s^{2}$

Karnataka CET-2022

Rotational Motion

149781 The angular speed of a fly wheel moving with uniform angular acceleration changes from $1200 rpm$ to $3120 rpm$ in 16 seconds. The angular acceleration in $rad / s^{2}$ is

1

12 π

2

104 π

3

2 π

4

4 π

Explanation:

D Given,
Angular acceleration changes from $1200 rpm$ to 3120 rpm in $16 \sec$ .
Angular acceleration
$α = \frac{ω_{f} - ω_{i}}{t} From equation ω_{f} = ω_{i} + α t]$
$ω_{f} = 3120 \times \frac{2 π}{60} rad / \sec$
$ω_{f} = 104 π rad / \sec$
$ω_{i} = 1200 \times \frac{2 π}{60} rad / \sec$
$ω_{i} = 40 π rad / \sec$
Then,
$α = \frac{104 π - 40 π}{16} = \frac{64 π}{16}$
$α = 4 π rad / \sec^{2}$

NEET-17.07.2022

Rotational Motion

149782 The angular velocity of a ceiling fan reduces to $50 %$ 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
$ω^{2} = ω_{o}^{2} - 2 α θ$
Angular velocity reduces $(50 %), ω = \frac{ω_{0}}{2}$
So, ${(\frac{ω_{0}}{2})}^{2} = ω_{0}^{2} - 2 α (2 π n) (\begin{matrix} ∵ θ = 2 π n \\ n = 36 \end{matrix})$
$α = \frac{3 ω_{0}^{2}}{4 \times 4 π \times 36}$
Now let Fan completes total $n$ revolution
From the starting to come to rest.
$0 = ω_{0}^{2} - 2 α (2 π n) \Rightarrow n = (\frac{ω_{0}^{2}}{4 α π})$
Substituting the value of $α$ from equation (i), we get-
$n = \frac{ω_{0}^{2}}{4 π} \times \frac{4 \times 4 π \times 36}{3 ω_{0}^{2}} = 48 revolution$
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 $100 π$ rad in the first 5 seconds. Its angular speed at the end of 5 seconds is

1

20 π rad . s

2

30 π rad . s^{- 1}

3

40 π rad . s^{- 1}

4

10 π rad . s^{- 1}

Explanation:

C Given, $ω_{i} = 0$ [start from rest]
$θ = 100 π rad$
$t = 5 \sec$
From using the second equation of angular of motion,
$θ = ω_{0} t + \frac{1}{2} α t^{2}$
$100 π = 0 + \frac{1}{2} \times α \times 25$
$α = \frac{200 π}{25}$
$α = 8 π rad / \sec^{2}$
Now, to find angular speed, we use -
$ω = ω_{0} + α t$
$ω = 0 + 8 π \times 5$
$ω = 40 π rad / \sec$

AP EAMCET-03.09.2021

Rotational Motion

149780 The angular speed of a motor wheel is increased from $1200 rpm$ to $3120 rpm$ in $16$ seconds. The angular acceleration of the motor wheel is

1

6 π rad / s^{2}

2

8 π rad / s^{2}

3

2 π rad / s^{2}

4

4 π rad / s^{2}

Explanation:

D Given,
$N_{1} = 1200 rpm$
$N_{2} = 3120 rpm$
$t = 16 \sec$
Initial angular speed-
$ω_{i} = \frac{2 π N_{1}}{60} = \frac{2 π \times 1200}{60} = 40 π rad / s$
Final angular speed-
$ω_{f} = \frac{2 π N_{2}}{60} = \frac{2 π \times 3120}{60} = 104 π rad / s$
So, Angular acceleration-
$α = \frac{(ω_{f} - ω_{i})}{t}$
$= \frac{(104 - 40) π}{16}$
$= 4 π rad / s^{2}$

Karnataka CET-2022

Rotational Motion

149781 The angular speed of a fly wheel moving with uniform angular acceleration changes from $1200 rpm$ to $3120 rpm$ in 16 seconds. The angular acceleration in $rad / s^{2}$ is

1

12 π

2

104 π

3

2 π

4

4 π

Explanation:

D Given,
Angular acceleration changes from $1200 rpm$ to 3120 rpm in $16 \sec$ .
Angular acceleration
$α = \frac{ω_{f} - ω_{i}}{t} From equation ω_{f} = ω_{i} + α t]$
$ω_{f} = 3120 \times \frac{2 π}{60} rad / \sec$
$ω_{f} = 104 π rad / \sec$
$ω_{i} = 1200 \times \frac{2 π}{60} rad / \sec$
$ω_{i} = 40 π rad / \sec$
Then,
$α = \frac{104 π - 40 π}{16} = \frac{64 π}{16}$
$α = 4 π rad / \sec^{2}$

NEET-17.07.2022

Rotational Motion

149782 The angular velocity of a ceiling fan reduces to $50 %$ 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
$ω^{2} = ω_{o}^{2} - 2 α θ$
Angular velocity reduces $(50 %), ω = \frac{ω_{0}}{2}$
So, ${(\frac{ω_{0}}{2})}^{2} = ω_{0}^{2} - 2 α (2 π n) (\begin{matrix} ∵ θ = 2 π n \\ n = 36 \end{matrix})$
$α = \frac{3 ω_{0}^{2}}{4 \times 4 π \times 36}$
Now let Fan completes total $n$ revolution
From the starting to come to rest.
$0 = ω_{0}^{2} - 2 α (2 π n) \Rightarrow n = (\frac{ω_{0}^{2}}{4 α π})$
Substituting the value of $α$ from equation (i), we get-
$n = \frac{ω_{0}^{2}}{4 π} \times \frac{4 \times 4 π \times 36}{3 ω_{0}^{2}} = 48 revolution$
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 $100 π$ rad in the first 5 seconds. Its angular speed at the end of 5 seconds is

1

20 π rad . s

2

30 π rad . s^{- 1}

3

40 π rad . s^{- 1}

4

10 π rad . s^{- 1}

Explanation:

C Given, $ω_{i} = 0$ [start from rest]
$θ = 100 π rad$
$t = 5 \sec$
From using the second equation of angular of motion,
$θ = ω_{0} t + \frac{1}{2} α t^{2}$
$100 π = 0 + \frac{1}{2} \times α \times 25$
$α = \frac{200 π}{25}$
$α = 8 π rad / \sec^{2}$
Now, to find angular speed, we use -
$ω = ω_{0} + α t$
$ω = 0 + 8 π \times 5$
$ω = 40 π rad / \sec$

AP EAMCET-03.09.2021