Suppose the rod in the previous problem has a mass of 1 kg distributed

1.	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» *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 :-)*

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»

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 :-)