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A car is moving in a circular horizontal track of radius 10 m with a constant speed of 10 m/s. A pendulum bob is suspended from the roof of the car by a light rigid rod of length 1.00 m. The angle made by the rod with track is
- a)zero
- b)30°
- c)45°
- d)60°
Correct answer is option 'C'. Can you explain this answer?
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A car is moving in a circular horizontal track of radius 10 m with a c...
When the car is moving in a circular horizontal track of radius 10 m with a constant speed, then the bob is also undergoing a circular motion. The bob is under the influence of two forces.
(i) T (tension in the rod)
(ii) mg (weight of the bob)
Resolving tension, we get
Tcosθ = mg ... (i)
And Tsinθ = mv2/r ... (ii)
(Here T sin θ is producing the necessary centripetal force for circular motion)
Dividing (ii) by (i), we get
Most Upvoted Answer
A car is moving in a circular horizontal track of radius 10 m with a c...
C) 45d) 60
The correct answer is c) 45 degrees.
To find the angle made by the rod with the track, we can use the concept of centripetal force.
The centripetal force is the force that keeps an object moving in a circular path. It is directed towards the center of the circle.
In this case, the centripetal force is provided by the tension in the rod. The tension in the rod can be resolved into two components: one along the direction of motion of the car, and the other perpendicular to it.
Since the car is moving with a constant speed in a circular path, the net force acting on it must be zero. This means that the perpendicular component of the tension must be equal in magnitude and opposite in direction to the gravitational force acting on the bob.
The gravitational force acting on the bob is given by the equation F = mg, where m is the mass of the bob and g is the acceleration due to gravity.
The tension in the rod can be calculated using the equation T = mg + ma, where a is the centripetal acceleration of the car.
The centripetal acceleration can be calculated using the equation a = v^2/r, where v is the speed of the car and r is the radius of the circular track.
Substituting the given values, we get a = (10 m/s)^2 / 10 m = 10 m/s^2.
Substituting the values of m, g, and a into the equation for tension, we get T = m(10 m/s^2) + mg.
Since the angle made by the rod with the track is the angle between the tension and the vertical direction, we can use trigonometry to find this angle.
The cosine of the angle is given by the equation cos(theta) = (mg)/T.
Substituting the values of m, g, and T into this equation, we get cos(theta) = (1 kg)(10 m/s^2)/(10 N) = 1/1 = 1.
Therefore, the angle made by the rod with the track is theta = cos^(-1)(1) = 45 degrees.
The correct answer is c) 45 degrees.
To find the angle made by the rod with the track, we can use the concept of centripetal force.
The centripetal force is the force that keeps an object moving in a circular path. It is directed towards the center of the circle.
In this case, the centripetal force is provided by the tension in the rod. The tension in the rod can be resolved into two components: one along the direction of motion of the car, and the other perpendicular to it.
Since the car is moving with a constant speed in a circular path, the net force acting on it must be zero. This means that the perpendicular component of the tension must be equal in magnitude and opposite in direction to the gravitational force acting on the bob.
The gravitational force acting on the bob is given by the equation F = mg, where m is the mass of the bob and g is the acceleration due to gravity.
The tension in the rod can be calculated using the equation T = mg + ma, where a is the centripetal acceleration of the car.
The centripetal acceleration can be calculated using the equation a = v^2/r, where v is the speed of the car and r is the radius of the circular track.
Substituting the given values, we get a = (10 m/s)^2 / 10 m = 10 m/s^2.
Substituting the values of m, g, and a into the equation for tension, we get T = m(10 m/s^2) + mg.
Since the angle made by the rod with the track is the angle between the tension and the vertical direction, we can use trigonometry to find this angle.
The cosine of the angle is given by the equation cos(theta) = (mg)/T.
Substituting the values of m, g, and T into this equation, we get cos(theta) = (1 kg)(10 m/s^2)/(10 N) = 1/1 = 1.
Therefore, the angle made by the rod with the track is theta = cos^(-1)(1) = 45 degrees.
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A car is moving in a circular horizontal track of radius 10 m with a constant speed of 10 m/s. A pendulum bob is suspended from the roof of the car by a light rigid rod of length 1.00 m. The angle made by the rod with track isa)zerob)30°c)45°d)60°Correct answer is option 'C'. Can you explain this answer?
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