Question

Guys...this may seems to be hard...bt no...it's not...I need to know the solution of only part:-(d) & (e)
(a)The moment of inertia of a disc of radius 0.5 m about its geometric axis is 2kg-m2. If a string is tied to
its circumference and a force of 10 Newton is applied, the value of torque with respect to this axis will be
(1) 2.5 N-m
21275 N-m
(3) 10 N-m
(4) 20 N-m
(b)In the above question, if the disc executes rotatory motion, its angular acceleration will be :-
(1) 2.5 rad/sec
(2) 5 rad/sec
(3) 10 rad/sec
(4) 20 rad/sec
(c)In the above question, the value of its angular velocity after 2 seconds will be :-
(1) 2.5 rad/sec
(2) 5 rad/sec
(3) 10 rad/sec
(4) 20 rad/sec
(d)In the above question, the change in angular momentum of disc in first 2 seconds in Nm second will be
(1) 2.5
(2) 5
(3) 10
(4) 20
(e)In the above question, angular displacement of the disc, in first two second will be in radian:-
(1) 2.5
(2) 5
(3) 10
(4) 20


Plz guys...I just don't wanna copy it in my hw...
I wanna learn how the correct answer came out...so plz explain your solution with proper formula and sign and don't spam...plz
Would be very thankful to you...!!!​

Answer 1

1. ( d)

(e) The angular displacement in the first 10 seconds is given by

θ=ω

0

t+

2

1

αt

2

=

2

1

(2.0rad/s

2

)(10s)

2

=100rad

As the wheel turns by 2π radian in each revolution, the number of revolutions in 10 s is

n=

2π

100

=16.

Explanation:

I hope it helps you.

Be happy