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Question: A bucket of water of mass \( 21kg \) is suspended by a rope wrapped around a solid cylinder \( 0.2{\...

A bucket of water of mass 21kg21kg is suspended by a rope wrapped around a solid cylinder 0.2 m0.2{\text{ m}} in diameter. The mass of the solid cylinder is 21kg21kg . The bucket is released from rest. Which of the following statements are correct?
(A) The tension in the rope is 70 N70{\text{ N}} .
(B) The acceleration of the bucket is 203m/s2\dfrac{{20}}{3}m/{s^2}
(C) The acceleration of the bucket is 403m/s2\dfrac{{40}}{3}m/{s^2}
(D) The tension in the rope is 140N140N

Explanation

Solution

Hint : As there are two unknowns here, so we will be using two equations, one will be obtained by balancing the forces on the bucket and the other will be obtained by the torque or moment equation for the solid cylinder around which the rope is wrapped.

Complete Step By Step Answer:
As the bucket will be experiencing the force due to gravitation which will be equal to,
f=m×gf = m \times g ,
Taking ‘g’ as 10ms210m{s^{ - 2}} and m is given as 21kg21kg , so the force will be,
f=21×10=210Nf = 21 \times 10 = 210N
This force will be pulling the bucket downward.
Now writing the equation for bucket,
F=ma\sum F = ma
210T=21a (1)210 - T = 21a{\text{ }} - - - - - - - - - (1)
Where T is the tension on the rope and let a is the acceleration of the bucket.
For calculating the tension on the rope let us take the moment equation around the solid cylinder into consideration,
τ=Iα\tau = I\alpha ,
Where, τ\tau is the torque about the solid cylinder and is given by T×rT \times r
I is the moment of inertia of the solid cylinder and is given by mr22\dfrac{{m{r^2}}}{2}
And α\alpha is the angular acceleration and is given by ar\dfrac{a}{r}
Now putting the above values in the equation of torque, we will get,
T×r=12mr2×arT \times r = \dfrac{1}{2}m{r^2} \times \dfrac{a}{r} ,
Simplifying this will give us,
T=ma2T = \dfrac{{ma}}{2}
T=21a2 - - - - - - - - - - - - (2)T = \dfrac{{21a}}{2}{\text{ - - - - - - - - - - - - (2)}}
Putting this in equation 11 ,
210ma2=21a210 - \dfrac{{ma}}{2} = 21a ,
As we know the value of m,
21021a2=21a210 - \dfrac{{21a}}{2} = 21a
Rearranging the above equation will give us,
210=21a+21a2210 = 21a + \dfrac{{21a}}{2}
On solving we will get, a=203ms2a = \dfrac{{20}}{3}m{s^{ - 2}}
Now putting this value of a in equation 22 ,
T=212×203T = \dfrac{{21}}{2} \times \dfrac{{20}}{3} ,
T=70NT = 70N
Hence, option A and B are the correct answers.

Note :
Though the exact value for the acceleration due to gravity is 9.8ms29.8m{s^{ - 2}} , but for convenience we have taken 10 to solve this question. It is a common practice and one should not get confused in the same. We can have a better understanding of the forces and their direction by making the FBD.