根据角动量定理,当物体受到外力时,其角动量的变化率等于外力的合力矩。因此,我们可以根据这个定理来计算停下转轮所需要的时间。 首先,我们需要计算转轮的角动量。由于转轮是刚体,其角动量可以用转动惯量和角速度的乘积来表示。转动惯量可以用半径和质量来计算,即: $I = \frac{1}{2}mr^2 = \frac{1}{2} \times 150\text{kg} \times (0.3\text{m})^2 = 13.5 \text{kg m}^2$ 其中,r为转轮的半径。 转轮的角速度为: $\omega = \frac{2\pi n}{60}= \frac{2\pi \times 1200}{60} = 125.66 \text{rad/s}$ 因此,转轮的角动量为: $L = I\omega = 13.5 \text{kg m}^2 \times 125.66 \text{rad/s} = 1698.3 \text{kg m}^2/s$ 现在,我们需要计算外力的合力矩。由于上下两个力的作用方向相反,因此两个力的合力矩为: $M = F_{up} \times r_{up} - F_{down} \times r_{down} = 2000 \text{N} \times 0.25 \text{m} - 1000 \text{N} \times 0.25 \text{m} = 250 \text{Nm}$ 根据角动量定理,外力的合力矩等于角动量的变化率。因此: $M = \frac{dL}{dt}$ 对上式两边同时积分,得到: $\int_0^t M dt = \int_{L_0}^0 dL$ 其中,$L_0$为转轮的初始角动量。将上面的数值代入,得到: $250 \text{Nm} \times t = 1698.3 \text{kg m}^2/s$ 解出t,得到: $t = \frac{1698.3 \text{kg m}^2/s}{250 \text{Nm}} \approx 6.79 \text{s}$ 因此,要停下转轮大约需要6.79秒的时间。用英文叙述
时间: 2023-07-25 16:24:57 浏览: 58
According to the law of conservation of angular momentum, the rate of change of angular momentum of an object is equal to the torque acting on it. Therefore, we can use this law to calculate the time required to stop the rotating wheel.
First, we need to calculate the angular momentum of the wheel. Since the wheel is a rigid body, its angular momentum can be expressed as the product of its moment of inertia and angular velocity. The moment of inertia can be calculated using the radius and mass of the wheel, which gives:
$I = \frac{1}{2}mr^2 = \frac{1}{2} \times 150\text{kg} \times (0.3\text{m})^2 = 13.5 \text{kg m}^2$
where r is the radius of the wheel.
The angular velocity of the wheel is:
$\omega = \frac{2\pi n}{60}= \frac{2\pi \times 1200}{60} = 125.66 \text{rad/s}$
Therefore, the angular momentum of the wheel is:
$L = I\omega = 13.5 \text{kg m}^2 \times 125.66 \text{rad/s} = 1698.3 \text{kg m}^2/s$
Now, we need to calculate the torque acting on the wheel. Since the two forces acting on the wheel are in opposite directions, the net torque is:
$M = F_{up} \times r_{up} - F_{down} \times r_{down} = 2000 \text{N} \times 0.25 \text{m} - 1000 \text{N} \times 0.25 \text{m} = 250 \text{Nm}$
According to the law of conservation of angular momentum, the torque acting on the wheel is equal to the rate of change of its angular momentum. Therefore:
$M = \frac{dL}{dt}$
Integrating both sides of the equation gives:
$\int_0^t M dt = \int_{L_0}^0 dL$
where $L_0$ is the initial angular momentum of the wheel. Substituting the values we calculated gives:
$250 \text{Nm} \times t = 1698.3 \text{kg m}^2/s$
Solving for t gives:
$t = \frac{1698.3 \text{kg m}^2/s}{250 \text{Nm}} \approx 6.79 \text{s}$
Therefore, it will take approximately 6.79 seconds to stop the rotating wheel.
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