Question

Suppose the rod in the previous problem has a mass of 1 kg distributed uniformly over its length,
(a) Find the initial angular acceleration of the rod ?
(b) Find the tension in the supports to the blocks of mass 2 kg and 5 kg ?

Answer 1

Given in the question :-

Mass of the rod = 1 kg.

(a)Now according to the formula we know that

$\tau_n_e_t = I_n_e_t + \alpha$

Since mass is unifirmly distributed hence there is no difference in the torque .

Now the moment of Inertia of the pivoted

=$ml^2/12$

= 1× 1² /12

= 1/12 kg.m²

Therefore the total MI will be.

= 1.75 + 1/12

= 1.75+0.083

=1.833 kg.m²

Now we have to find the angular acceleration i.e α

$\alpha = T/I$

Put the given values

=1.5× 9.8/1.833

=8.0 rad/ s²

(b) Now for finding the Linear acceleration

$a = \alpha r$

=8.0 × 0.5

=4.0 m /s²

Now weight 2g N is on the downwards direction , hence in result tension t and accelration a will be in upward.

Now,

From the given condition at 2kg mass.

$T- 2g = 2a$

T=2g+2a

=2× 9.8+2 × 4.0

=19.6+8

= 27.6 N

Hence accleration is in downwards direction at 5 kg mass.

$5 g-T =5 a$

T=5 g-5 a

=5 × 9.8 -5 × 4

=49-20

=29 N.

Hope it Helps :-)

Answer 2

Explanation:

Given in the question :-

Mass of the rod = 1 kg.

(a)Now according to the formula we know that

\tau_n_e_t = I_n_e_t + \alpha

Since mass is unifirmly distributed hence there is no difference in the torque .

Now the moment of Inertia of the pivoted

=ml^2/12ml

2

/12

= 1× 1² /12

= 1/12 kg.m²

Therefore the total MI will be.

= 1.75 + 1/12

= 1.75+0.083

=1.833 kg.m²

Now we have to find the angular acceleration i.e α

\alpha = T/Iα=T/I

Put the given values

=1.5× 9.8/1.833

=8.0 rad/ s²