1. (a) In a uniform circular motion will there be any acceleration?Explain your answer. (2 marks) (b) If A=3hat (i)-hat (j)+2hat (k) and B=3hat (i)+3hat (j)-2hat (k) find A.B (2 marks) (c) A force field is given by the potential U=x^2+2y^2+z^3+2xy+2yz+zx Find the work done by the force field in moving a particle from the point (1,1,2) to (2,1,2) (3 marks) (d) (i) Show that in a simple pendulum when the bob is given a small displacement , it makes a simple harmonic motion. (5 marks) (ii) Determine the length of a simple pendulum that will swing back and forth in simple harmonic motion with a period of 2 seconds (Use the value of acceleration due to gravity g=10.0m/s^2 (3 marks) (e) Find, from the first principle, the moment of inertia of a thin rod of length /about an axis through its centre and perpendicular to the length of the rod. (5 marks)

1. (a) Yes, there is acceleration in uniform circular motion. In uniform circular motion, the speed of the object remains constant, but its direction continuously changes as it moves along the circular path. This change in direction means that there is a centripetal acceleration directed towards the center of the circle.<br /><br />(b) To find the dot product \( A \cdot B \), we use the formula:<br />\[ A \cdot B = (3\hat{i} - \hat{j} + 2\hat{k}) \cdot (3\hat{i} + 3\hat{j} - 2\hat{k}) \]<br />\[ = (3)(3) + (-1)(3) + (2)(-2) \]<br />\[ = 9 - 3 - 4 \]<br />\[ = 2 \]<br /><br />(c) The work done by a force field given by a potential \( U \) when moving a particle from point \( (x_1, y_1, z_1) \) to \( (x_2, y_2, z_2) \) is given by the change in potential energy:<br />\[ W = U(x_1, y_1, z_1) - U(x_2, y_2, z_2) \]<br />First, calculate \( U(1, 1, 2) \):<br />\[ U(1, 1, 2) = 1^2 + 2(1)^2 + 2^3 + 2(1)(1) + 2(1)(2) + (2)(1) \]<br />\[ = 1 + 2 + 8 + 2 + 4 + 2 \]<br />\[ = 19 \]<br /><br />Now, calculate \( U(2, 1, 2) \):<br />\[ U(2, 1, 2) = 2^2 + 2(1)^2 + 2^3 + 2(2)(1) + 2(1)(2) + (2)(2) \]<br />\[ = 4 + 2 + 8 + 4 + 4 + 4 \]<br />\[ = 26 \]<br /><br />The work done is:<br />\[ W = 19 - 26 = -7 \]<br /><br />(d) (i) For a simple pendulum with small displacement, the restoring force is proportional to the sine of the angle of displacement. For small angles, \(\sin \theta \approx \theta\), making the restoring force approximately proportional to the displacement. This results in simple harmonic motion, where the acceleration is directly proportional to the negative of the displacement.<br /><br />(ii) The period \( T \) of a simple pendulum is given by:<br />\[ T = 2\pi \sqrt{\frac{L}{g}} \]<br />Given \( T = 2 \) seconds and \( g = 10.0 \, \text{m/s}^2 \), solve for \( L \):<br />\[ 2 = 2\pi \sqrt{\frac{L}{10}} \]<br />\[ 1 = \pi \sqrt{\frac{L}{10}} \]<br />\[ \frac{1}{\pi} = \sqrt{\frac{L}{10}} \]<br />\[ \left(\frac{1}{\pi}\right)^2 = \frac{L}{10} \]<br />\[ L = 10 \left(\frac{1}{\pi}\right)^2 \]<br />\[ L \approx \frac{10}{\pi^2} \approx 1.013 \, \text{m} \]<br /><br />(e) To find the moment of inertia \( I \) of a thin rod of length \( l \) about an axis through its center and perpendicular to its length, consider an infinitesimal element of mass \( dm \) at a distance \( x \) from the center. The linear mass density is \( \lambda = \frac{M}{l} \), where \( M \) is the total mass of the rod. The moment of inertia is given by:<br />\[ I = \int_{-\frac{l}{2}}^{\frac{l}{2}} x^2 \, dm \]<br />Since \( dm = \lambda \, dx = \frac{M}{l} \, dx \), we have:<br />\[ I = \int_{-\frac{l}{2}}^{\frac{l}{2}} x^2 \frac{M}{l} \, dx \]<br />\[ = \frac{M}{l} \int_{-\frac{l}{2}}^{\frac{l}{2}} x^2 \, dx \]<br />\[ = \frac{M}{l} \left[ \frac{x^3}{3} \right]_{-\frac{l}{2}}^{\frac{l}{2}} \]<br />\[ = \frac{M}{l} \left( \frac{\left(\frac{l}{2}\right)^3}{3} - \frac{\left(-\frac{l}{2}\right)^3}{3} \right) \]<br />\[ = \frac{M}{l} \left( \frac{l^3}{24} + \frac{l^3}{24} \right) \]<br />\[ = \frac{M}{l} \cdot \frac{l^3}{12} \]<br />\[ = \frac{Ml^2}{12} \]