Question

A horizontal 745 N merry-go-round of radius

Answer 1

Answer:

The kinetic energy of the merry-goround after 3.62 s is  544J

Explanation:

Given :

Weight w = 745 N

Radius r =  1.45 m

Force =  56.3 N

To Find:

The kinetic energy of the merry-go round after 3.62  = ?

Solution:

Step 1:  Finding the Mass of merry-go-round

$m = \frac{ weight}{g}$

$m = \frac{745}{9.81 }$

m = 76.02 kg

Step 2: Finding the Moment of Inertia of solid cylinder

Moment of Inertia of solid cylinder I =$0.5 \times m \times r^2$

Substituting the values

Moment of Inertia of solid cylinder I  

=>$0.5 \times 76.02 \times (1.45)^2$

=> $0.5 \times 76.02\times 2.1025$

=> $79.91 kg.m^2$

Step 3: Finding the Torque applied T

Torque applied T = $F \times r$

Substituting the values

T = $56.3 \times 1.45$

T = 81.635 N.m

 Step 4: Finding the Angular acceleration

Angular acceleration ,$\alpha = \frac{Torque}{Inertia}$

Substituting the values,

$\alpha = \frac{81.635}{79.91}$

$\alpha = 1.021 rad/s^2$

 Step 4: Finding the Final angular velocity

Final angular velocity ,$\omega = \alpha \times t$

Substituting the values,

$\omega = 1.021 \times 3.62$

$\omega = 3.69 rad/s$

Now KE (100% rotational) after 3.62s is:

KE = $0.5 \times I \times \omega^2$

KE =$0.5 \times 79.91 \times 3.69^2$

KE = 544J