1.(a) A solid sphere of mass 100gm and radius 2.5cm rolls without sliding wit ha uniform velocity of 10cm/sec along a straight line on a smooth horizontal table. Calculate the total energy.
(b) A hoop of radius 100cm and mass 19kg is rolling along a horizontal surface, so that its center of mass has a velocity of 20cm/sec. How much work will have to be done to stop it?
2 answers
Subject is not CUC but physics
In each case, add the translational kinetic energy, (M/2)V^2, to the rotational kinetic energy, (I/2)w^2.
I is the moment of intertia about the center of mass and w is the angular velocity.
(1a) (M/2)V^2 = (1/2)*0.050*(0.1)^2
= 2.5*10^-4 J
(I/2)w^2 = (1/2)(2/5)M*R^2*(V/R)^2
= (M/5)V^2 = 1.0*10^-4 J
Total = 3.5*10^-4 J
(1b) (M/2)V^2 = (0.5)(0.019)(0.2)^2 J
(I/2)w^2 = (0.5)M*R^2(V/R)^2
= (0.5)M*V^2
Translational and rotational KE are equalo for the hoop. The work required to stop is the negative of that.
I is the moment of intertia about the center of mass and w is the angular velocity.
(1a) (M/2)V^2 = (1/2)*0.050*(0.1)^2
= 2.5*10^-4 J
(I/2)w^2 = (1/2)(2/5)M*R^2*(V/R)^2
= (M/5)V^2 = 1.0*10^-4 J
Total = 3.5*10^-4 J
(1b) (M/2)V^2 = (0.5)(0.019)(0.2)^2 J
(I/2)w^2 = (0.5)M*R^2(V/R)^2
= (0.5)M*V^2
Translational and rotational KE are equalo for the hoop. The work required to stop is the negative of that.
7 answers