269589 A thin metal rod of length \(0.5 \mathrm{~m}\) is vertically straight on horizontal floor. This rod is falling freely to a side without slipping. The angular velocity of rod when its top end touches the floor is (nearly)

269590 What should be the minimum coefficient of static friction between the plane and the cylinder, for the cylinder not to slip on an inclined plane

269591 A thin metal disc of radius \(25 \mathrm{~cm}\) and mass \(2 \mathrm{~kg}\) starts from rest and rolls down on an inclined plane. If its rotational kinetic energy is \(8 \mathrm{~J}\) at the foot of this inclined plane, then linear velocity of centre of mass of disc is

269648 Assume the earth's orbit around the sun as circular and the distance between their centres as ' \(D\) ' Mass of the earth is ' \(M\) ' and its radius is ' \(R\) ' If earth has an angular velocity ' \(\omega \omega_{0}\) ' with respect to its centre and ' \(\omega\) ' with respect to the centre of the sun, the total kinetic energy of the earth is:

1

2

3

4 \(\frac{2}{5} M R^{2} \omega_{0}^{2} \frac{\square}{\square}+\square \omega \Pi_{0}+\frac{5}{2} \square \frac{D \omega}{R \omega_{0}} \overbrace{}^{2}]\)

269589 A thin metal rod of length \(0.5 \mathrm{~m}\) is vertically straight on horizontal floor. This rod is falling freely to a side without slipping. The angular velocity of rod when its top end touches the floor is (nearly)

269590 What should be the minimum coefficient of static friction between the plane and the cylinder, for the cylinder not to slip on an inclined plane

269591 A thin metal disc of radius \(25 \mathrm{~cm}\) and mass \(2 \mathrm{~kg}\) starts from rest and rolls down on an inclined plane. If its rotational kinetic energy is \(8 \mathrm{~J}\) at the foot of this inclined plane, then linear velocity of centre of mass of disc is

269648 Assume the earth's orbit around the sun as circular and the distance between their centres as ' \(D\) ' Mass of the earth is ' \(M\) ' and its radius is ' \(R\) ' If earth has an angular velocity ' \(\omega \omega_{0}\) ' with respect to its centre and ' \(\omega\) ' with respect to the centre of the sun, the total kinetic energy of the earth is:

1

2

3

4 \(\frac{2}{5} M R^{2} \omega_{0}^{2} \frac{\square}{\square}+\square \omega \Pi_{0}+\frac{5}{2} \square \frac{D \omega}{R \omega_{0}} \overbrace{}^{2}]\)

269589 A thin metal rod of length \(0.5 \mathrm{~m}\) is vertically straight on horizontal floor. This rod is falling freely to a side without slipping. The angular velocity of rod when its top end touches the floor is (nearly)

269590 What should be the minimum coefficient of static friction between the plane and the cylinder, for the cylinder not to slip on an inclined plane

269591 A thin metal disc of radius \(25 \mathrm{~cm}\) and mass \(2 \mathrm{~kg}\) starts from rest and rolls down on an inclined plane. If its rotational kinetic energy is \(8 \mathrm{~J}\) at the foot of this inclined plane, then linear velocity of centre of mass of disc is

269648 Assume the earth's orbit around the sun as circular and the distance between their centres as ' \(D\) ' Mass of the earth is ' \(M\) ' and its radius is ' \(R\) ' If earth has an angular velocity ' \(\omega \omega_{0}\) ' with respect to its centre and ' \(\omega\) ' with respect to the centre of the sun, the total kinetic energy of the earth is:

1

2

3

4 \(\frac{2}{5} M R^{2} \omega_{0}^{2} \frac{\square}{\square}+\square \omega \Pi_{0}+\frac{5}{2} \square \frac{D \omega}{R \omega_{0}} \overbrace{}^{2}]\)

269589 A thin metal rod of length \(0.5 \mathrm{~m}\) is vertically straight on horizontal floor. This rod is falling freely to a side without slipping. The angular velocity of rod when its top end touches the floor is (nearly)

269590 What should be the minimum coefficient of static friction between the plane and the cylinder, for the cylinder not to slip on an inclined plane

269591 A thin metal disc of radius \(25 \mathrm{~cm}\) and mass \(2 \mathrm{~kg}\) starts from rest and rolls down on an inclined plane. If its rotational kinetic energy is \(8 \mathrm{~J}\) at the foot of this inclined plane, then linear velocity of centre of mass of disc is

269648 Assume the earth's orbit around the sun as circular and the distance between their centres as ' \(D\) ' Mass of the earth is ' \(M\) ' and its radius is ' \(R\) ' If earth has an angular velocity ' \(\omega \omega_{0}\) ' with respect to its centre and ' \(\omega\) ' with respect to the centre of the sun, the total kinetic energy of the earth is:

1

2

3

4 \(\frac{2}{5} M R^{2} \omega_{0}^{2} \frac{\square}{\square}+\square \omega \Pi_{0}+\frac{5}{2} \square \frac{D \omega}{R \omega_{0}} \overbrace{}^{2}]\)