Respuesta :
Answer:
The value of the constant force is [tex]\bf{296.88~N}[/tex].
Explanation:
Given:
Mass of the merry-go-round, [tex]m = 210~Kg[/tex]
Radius of the horizontal disk, [tex]r = 1.5~m[/tex]
Time required, [tex]t = 2.00~s[/tex]
Angular speed, [tex]\omega = 0.600~rev/s[/tex]
Torque on an object is given by
[tex]\tau = F.r = I.\alpha~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)[/tex]
where [tex]I[/tex] is the moment of inertia of the object, [tex]\alpha[/tex] is the angular acceleration and [tex]F[/tex] is the force on the disk.
The moment of inertia of the horizontal disk is given by
[tex]I = \dfrac{1}{2}mr^{2}[/tex]
and the angular acceleration is given by
[tex]\alpha = \dfrac{2\pi \omega}{t}[/tex]
Substituting all these values in equation (1), we have
[tex]F &=& \dfrac{I\alpha}{r}\\&=& \dfrac{\pi m r \omega}{t}\\&=& \dfarc{\pi(210~Kg)(1.5~m)(0.600~rev/s)}{2.00~s}\\&=& 296.88~N[/tex]
