Asked by Theresa
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?
(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?
Answers
Answered by
Theresa
Subject is not CUC but physics
Answered by
drwls
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.
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