**1. A particle moves in a circle of radius 20 cm at a speed that increases uniformly. If the speed changes from 5 m/s to 6 m/s in 2 s, find the angular acceleration.**

It is given that speed of the particle increases uniformly which means the rate of change of speed is constant (with position or time). Since

*magnitude of tangential acceleration*is nothing but the rate of change of linear speed, it is also constant here which again implies that the average and instantaneous values of the same are equal.

Therefore the angular acceleration of the particle is 2.5 rad/s².

**2. What will be the magnitude of the acceleration of a particle traveling in a circle of radius 10 cm with uniform speed completing the circle in 4 s?**

**The particle moves in the circle with a constant speed implying that the tangential component of the resultant acceleration is zero. The net acceleration vector is always directed towards the centre or only the centripetal acceleration is present which needs to be evaluated.**

It is clear from the equation one that to get the acceleration we first need the speed v. Time Period of the circular motion is given to be 4s. We can thus write

Therefore the magnitude of the resultant acceleration vector is 2.5 π² cm/s² and it is always directed towards the centre of the circular path.

**3. A fan moves with angular velocity of 10π rad/s. Now it is switched off and an angular retardation of 2π rad/s² is produced. Find the number of rotations made by the fan before it stops.**

**The angular retardation produced is constant and therefore the basic equations of motion (in terms of axial variables) are valid here. The provided information regarding the fan’s motion is**

So the answer is 12.5 rotations.

**4. A particle starts moving in a circle of radius 1m with time dependent angular acceleration α = 2t² (α and t are in SI units). Find the magnitude of resultant acceleration at time t = 1s.**

The particle's acceleration will have two rectangular components, one along the radial direction (known as radial or centripetal acceleration) and the other directed tangentially (i.e. the tangential acceleration). The two individual components should be evaluated first and then they can be combined vectorially to obtain the resultant.

The relation between angular acceleration and tangential acceleration for circular motion is α = aₜ/R. Also α as a function of time t is known so aₜ at general instant t is also known (using the above cited relation). The radial acceleration aₙ can be found by using the relation aₙ = ω²R but for this we first require to get ω(t) from the provided function α(t).

Hence the magnitude of resultant acceleration at time t = 1s is 2.05 m/s².

**5. The angular velocity (ω) of a particle depends on its angular position (Ɵ, measured with respect to a certain line of reference) by the rule ω = 2√Ɵ. Find the angular acceleration α as a function of Ɵ.**

**The given function is**ω = 2√Ɵ. The problem asks to find the dependency of angular acceleration α over the angular position Ɵ. Angular acceleration is defined as the time rate of change of angular velocity. Therefore we can write the following:

Hence the angular acceleration in this case is a constant value 2 rad/s² and it is independent of the angular position Ɵ.