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Phoenix [80]
3 years ago
12

Quizlet The Cajun Cliffhanger at Great AMerica is a ride in which occupants line the perimeter of a cylinder and spin in a circl

e at a high rate of turning. When the cylinder begins spinning very rapidly, the floor is removed from under the rider's feet. What affect does a doubling in speed have upon the centripetal force?
Physics
1 answer:
Dmitrij [34]3 years ago
7 0

Answer:

Centripetal Force is the force that keeps the object's rotation in a circular path and it's always directed towards the center of rotation. Mathematically, it is given as,

F_{c}  = \frac{mv^{2} }{r}

From this equation it can be concluded that the centripetal force is proportional to the product of mass and the square of velocity. It is inversely proportional to the object's distance from the center.

Now in the above mentioned case, when the cylinder begins spinning very rapidly and it's speed becomes double. Then by applying the above equation we conclude that " the force will be increased four times if the speed is doubled as it is proportional to square of speed."

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a 45kg boy is holding a 12kg pumpkin while standing on ice skates on a smooth frozen pond. the boy tosses the pumpkin with a hor
irga5000 [103]
Let's split the problem into two parts, and let's solve it by using conservation of momentum.

1) Let's analyze the moment when the boy tosses the pumpkin. Before this moment, the boy is standing with the pumpkin, so with speed v=0 and therefore the total momentum of the system (boy+pumpkin) is zero.
After he tosses the pumpkin, the boy will start to move with speed v_b and the pumpkin starts to move with speed v_p=2.8 m/s. The momentum of the boy is m_b v_b (with m_b =45 kg being the mass of the boy), while the momentum of the pumpkin is m_p v_p (where m_p=12 kg is the mass of the pumpkin). Since the total momentum must be equal to zero (because the total momentum cannot change), then we can write
m_b v_b + m_p v_p =0
From which we find
v_b = - \frac{m_p v_p}{m_b}=- \frac{(12 kg)(2.8 m/s)}{45 kg} =-0.75 m/s
and this is the speed of the boy, and the negative sign means he's moving in the opposite direction of the pumpkin.

2) Now let's focus on the entire system (boy+pumpkin+girl). Initially, the total momentum of this system is zero, because both the boy (holding the pumpkin) and the girl are still. So, the total momentum after the girl catches the pumpkin must be still zero.
After this moment, the boy has a momentum of m_b v_b, while the girl has momentum (m_g+m_p)v_g where m_g=37 kg is the girl mass v_g is the girl speed. Here we use m_g+m_p because the girl is holding the pumpkin now. Therefore, the conservation of momentum becomes
m_bv_b + (m_g+m_p)v_g =0
and so
v_g = - \frac{m_b v_b}{m_g+m_p} =- \frac{(45 kg)(-0.75 m/s)}{37 kg+12 kg} =0.69 m/s
and this is the girl's speed, with positive sign so with same direction of the pumpkin initial direction.
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