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A merry- go- round ( diameter 4.0 m ) starts from rest and uniformly accelerates until it rotates at 0.8 rev/min. This occurs in 11s. Once it has reached this rate , it rotates at 0.8 rev/min uniformly . Determine the angular acceleration
  • a)
    7.55 x 10-3 rad/s2​​
  • b)
    6.7 x 10-3 rad/s2
  • c)
    0.083 rad/s2
  • d)
    0.083 rad/s
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
A merry- go- round ( diameter 4.0 m ) starts from rest and uniformly a...
First we have to find ‘v’ by using the relation v = rw
Now, v = 0.167m/s
Then we find ‘a’ by using the relation a = v/t
So, a = 0.0152m/s2.
Now, we can find angular acceleration by using the formula α = a/r
Finally, α = 0.0076 rad/sec2
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Community Answer
A merry- go- round ( diameter 4.0 m ) starts from rest and uniformly a...
Given data:
Diameter of the merry-go-round, d = 4.0 m
Time taken to reach the final speed, t = 11 s
Final speed, ω = 0.8 rev/min

We need to find the angular acceleration, α.

Formula used:
The angular acceleration, α is given by:
α = (ωf - ωi) / t

where,
ωi = initial angular velocity
ωf = final angular velocity
t = time taken to change the velocity

Conversion of units:
ω = 0.8 rev/min = (0.8 × 2π) rad/min = 1.6π rad/min
t = 11 s

Calculation:
The diameter of the merry-go-round is 4.0 m, hence the radius is r = 2.0 m.
The initial angular velocity, ωi can be calculated using the formula:
v = ωir

where,
v = velocity
r = radius

Initially, the merry-go-round was at rest, so the initial velocity, vi = 0.
Hence, ωi = vi/r = 0/r = 0

Substituting the values in the formula for angular acceleration, we get:
α = (ωf - ωi) / t
α = (1.6π - 0) / 11
α = 0.145 radians/s^2

Therefore, the correct option is (A) 7.55 x 10^-3 rad/s^2.
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A merry- go- round ( diameter 4.0 m ) starts from rest and uniformly accelerates until it rotates at 0.8 rev/min. This occurs in 11s. Once it has reached this rate , it rotates at 0.8 rev/min uniformly . Determine the angular accelerationa)7.55 x 10-3rad/s2​​b)6.7 x 10-3rad/s2c)0.083 rad/s2d)0.083 rad/sCorrect answer is option 'A'. Can you explain this answer? for Class 11 2026 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about A merry- go- round ( diameter 4.0 m ) starts from rest and uniformly accelerates until it rotates at 0.8 rev/min. This occurs in 11s. Once it has reached this rate , it rotates at 0.8 rev/min uniformly . Determine the angular accelerationa)7.55 x 10-3rad/s2​​b)6.7 x 10-3rad/s2c)0.083 rad/s2d)0.083 rad/sCorrect answer is option 'A'. Can you explain this answer? covers all topics & solutions for Class 11 2026 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A merry- go- round ( diameter 4.0 m ) starts from rest and uniformly accelerates until it rotates at 0.8 rev/min. This occurs in 11s. Once it has reached this rate , it rotates at 0.8 rev/min uniformly . Determine the angular accelerationa)7.55 x 10-3rad/s2​​b)6.7 x 10-3rad/s2c)0.083 rad/s2d)0.083 rad/sCorrect answer is option 'A'. Can you explain this answer?.
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