A car with fuzzy dice hanging from the rear view mirror by a single string drives along at a
constant 25.0 m/s. a) If the dice have a mass of 0.030 kg, what is the angle of the string (relative to
vertical)? b) What is the tension in the string? c) As the driver takes a turn with a radius of 150.0 m,
what will the angle then be? (at same speed) d) What is the tension in the string during this turn?
(a) v=const, a=0 =>α =0,
(b)T=m•g.
(c)
m•v² /R=T•sin α,
m•g=T•cos α.
m•v²/R•m•g=T•sin α/T• cos α.
v²/R •g = tan α.
α =arctan(v²/R •g).
(d)
T=mg/cosα
To answer these questions, we'll need to use some basic principles of physics, including Newton's laws of motion and centripetal force.
a) To determine the angle of the string relative to the vertical, we need to consider the forces acting on the dice. Since the car is moving at a constant velocity, we know that the net force acting on the dice in the vertical direction must be zero. The forces acting on the dice are its weight (mg) in the downward direction and the tension force (T) in the upward direction. The angle of the string can be determined using trigonometry.
Let's assume θ represents the angle of the string relative to the vertical. Using trigonometry, we know that sin(θ) = opposite/hypotenuse, which in this case is the weight (mg) divided by the tension force (T). Therefore, sin(θ) = mg/T.
b) To find the tension in the string, we'll solve the equation from part a) for T. Rearranging the equation, we get T = mg/sin(θ). Plugging in the values, T = (0.030 kg)(9.8 m/s²) / sin(θ).
c) When the car takes the turn, we introduce a new force called the centripetal force. This force acts towards the center of the turn and is necessary to keep the car moving in a circular path. The angle of the string will change due to the change in direction caused by the centripetal force.
To calculate the new angle, we consider the car in the turn with the radius of 150.0 m. The force equation for circular motion is Fc = mv²/r, where Fc is the centripetal force, m is the mass of the dice, v is the velocity of the car, and r is the radius of the turn.
Since the car is moving at a constant velocity, the net force in the horizontal direction must be zero. The horizontal forces acting on the dice are the tension force (T) and the centripetal force (Fc). The angle of the string can be determined using trigonometry.
Let's assume θ' represents the new angle of the string. Using trigonometry again, we know that sin(θ') = opposite/hypotenuse, which in this case is the centripetal force (Fc) divided by the tension force (T). Therefore, sin(θ') = Fc/T.
d) To find the tension in the string during the turn, we'll solve the equation from part c) for T. Rearranging the equation, we get T = Fc/sin(θ'). Plugging in the values, T = (m v²)/r / sin(θ').
This is how you can determine the angle of the string and the tension in the string for both the constant velocity and during the turn in the car.