Question

1. a bucket of water of mass 5 kg is tied with a rope of length 50cm is rotated in a vertical circle.if the bucket rotates 10 rotations i 1 min,calculate the maximum velocity with which it can be rotated without spilling the water at the highest point

2. the centripetal force acting on a body of mas 500g is 10x10 power of -2 Newton.if the radius of the path is 10m,calculate the velocity with which the body moves

3. two forces of magnitude 5N and 10N are incligned with each other by 30 degree.what is the magnitud and direction of the resultant

4. a cyclist travel in a circular arbit of radius 200m.the angle of baking is 30 degree,what is the maximum velocity with which he can travel without skidding

5. a car of weight 1000kg travel with a speed of 25km/h in a curved path of radius 100m.what is the angle through which the outer track is raised so as to have a safe turn

Answer 1

Answer on Question #44866-Physics-Mechanics-Kinematics-Dynamics

1. A bucket of water of mass $5\mathrm{kg}$ is tied with a rope of length $50\mathrm{cm}$ is rotated in a vertical circle. If the bucket rotates 10 rotations i 1 min, calculate the maximum velocity with which it can be rotated without spilling the water at the highest point

Solution

$$m g = \frac {m v ^ {2}}{r} \rightarrow v = \sqrt {g r} = \sqrt {0 . 5 0 \cdot 9 . 8} = 2. 2 \frac {m}{s}.$$

2. The centripetal force acting on a body of mass $m = 500g = 0.5kg$ is $F = 10 \cdot 10^{-2}$ Newton = 0.1N. If the radius of the path is 10m, calculate the velocity with which the body moves

Solution

$$F = \frac {m v ^ {2}}{r} \rightarrow v = \sqrt {\frac {F r}{m}} = \sqrt {\frac {0 . 1 \cdot 1 0}{0 . 5}} = \sqrt {2} \frac {m}{s}.$$

3. Two forces of magnitude 5N and 10N are inclined with each other by 30 degree. What is the magnitude and direction of the resultant?

Solution

$$R ^ {2} = 5 ^ {2} + 1 0 ^ {2} + 2 \cdot 5 \cdot 1 0 \cos 3 0 = 1 2 5 + 5 0 \sqrt {3} \rightarrow R = 2 1 1. 6 N$$

Direction

$$\tan \alpha = \frac {Q \sin 3 0}{P + Q \cos 3 0} = \frac {1 0 \cdot 0 . 5}{5 + \frac {1 0 \sqrt {3}}{2}} \rightarrow \alpha = 2 0. 1 {}^ {\circ}.$$

The resultant makes $\alpha = 20.1{}^{\circ}$ with the line of force $5N$, towards the $10N$ force.

4. A cyclist travel in a circular orbit of radius 200m. the angle of baking is 30 degree, what is the maximum velocity with which he can travel without skidding

Solution

$N$ is normal force from ground supplies centripetal force and balances weight, $m$ is mass of vehicle and person.

$$N \cos 3 0 = m g \rightarrow N = \frac {2 m g}{\sqrt {3}}$$ $$N \sin 3 0 = \frac {m v ^ {2}}{R} \rightarrow v ^ {2} = \frac {2 m g}{2 \sqrt {3}} \cdot \frac {R}{m} = \frac {R g}{\sqrt {3}} = \frac {2 0 0 \cdot 9 . 8}{\sqrt {3}}.$$ $$v = 3 3. 6 \frac {m}{s}.$$

5. A car of weight $1000\mathrm{kg}$ travel with a speed of $25\mathrm{km/h}$ in a curved path of radius $100\mathrm{m}$. what is the angle through which the outer track is raised so as to have a safe turn

Solution

$$N \sin \theta = \frac{m v^{2}}{r}.$$ $$N \cos \theta = m g.$$ $$\tan \theta = \frac{\frac{m v^{2}}{r}}{m g} = \frac{v^{2}}{r g}. \theta = \tan^{-1} \frac{v^{2}}{r g} = \tan^{-1} \frac{\left(\frac{25 \, \text{m}}{3.6 \, \text{s}}\right)^{2}}{100 \, \text{m} \cdot 9.8 \, \frac{\text{m}}{\text{s}^{2}}} = 2.8{}^{\circ}.$$

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